Study Barn

A. Electric.

B. Heat.

C. Light.

D. All of the above.

Solar energy is used for all of the above options.

A. Nuclear.

B. Petroleum.

C. Wind.

D. Coal.

The United States has a large supply of coal. On top of that an infrastructure that has used coal for energy for decades. Thus coal is the most used source of energy in the US.

A. Heat.

B. Food.

C. Electricity.

D. Nuclear.

Most of the energy in our bodies is spent producing heat.

A. Biomass.

B. Coal.

C. Wind.

D. Solar.

Coal is not a renewable energy as it is burned in order to produce energy. Once it is burned it cannot be turned into more coal.

A. Energy is always in motion.

B. Energy is saved.

C. Energy is recycled.

D. Energy is neither created nor destroyed.

The law of conservation of energy states that energy may neither be created nor destroyed. Therefore the sum of all the energies in the system is a constant.

A. Nuclear energy.

B. Chemical energy.

C. Gravitational energy.

D. Kinetic energy.

Gravitational energy, Chemical energy, and Nuclear energy are all forms of potential energy. Kinetic energy is energy in motion and thus the opposite of potential energy.

A. electricity in action.

B. electricity that is stored.

C. stored energy.

D. work that has not been performed.

Potential energy, or stored energy, is the ability of a system to do work due to its position or internal structure.

A. reflection of light.

B. ability to rest.

C. ability to do work.

D. an electric force.

Energy is the amount of work that can be performed by a force.

A. (x)•(1) = x

B. x + 0 = x

C. x - x = 0

D. (x)•(1/x) = 1

E. (x)•(1) = y

Zero is the additive identity. Adding zero to any number will give an answer equal to the original number.

A. 1

B. 2

C. 3

D. 4

E. 5

(x/2) = (x/4) + (1/2)
Find a common denominator.
(2x/4) = (x/4) + (2/4)
Multiply each side by 4.
2x = x + 2
Subtract x from each side.
x = 2

A. 5n + 10

B. 5n + 2

C. n + 10

D. 10n

E. 5n

A regular pentagon has 5 sides of equal length. The perimeter would be 5(n+2) which equals 5n+10.

A. 1/3 ÷ 1/4

B. 1/4 ÷ 1/3

C. 1 - (1/3 + 1/4)

D. 1 + (1/3 + 1/4)

E. 1 + (2/3 + 1/4)

This part and whole problem opens up when you get the numbers out of the words.
1 day - (1/3 +1/4)

A. 39.2

B. 78.5

C. 314.2

D. 476.4

E. 628.4

Be Careful! This problem give the diameter (10), but the area formula requires the radius (5). Area of a circle is pi times the radius squared.

A. -100

B. -4

C. 4

D. 24

E. 1600

Quotient means to divide. Remember that a negative number divided by a negative number will give a positive answer.

A. 6

B. 8

C. 16

D. 24

E. 36

Draw a Venn diagram (circles) to represent the students taking art and music. The circles will overlap. The circle for art contains 26 students. The circle for music contains 18 students. Since 2 students take BOTH art and music (place 2 in the area where the circles overlap), the remainder of the art circle contains 24 students and the remainder of the music circle contains 16 students. Adding all values (24 + 16 + 2) gives a total of 42 students. Of the class of 50 students, 8 do not take art or music.

A. None.

B. I only.

C. I and II only.

D. I and III only.

E. I, II, and III.

Take control -- put in your own numbers. This is a COULD question, so you will want to try several different numbers to attempt to make I, II, and III true.
The exponents might have tipped you off that a negative fraction would be a good choice. Putting in –1/2 for n we find:
I. 2n = 2(–1/2) = –1 < n² =(–1/2)² = 1/4. TRUE
II. 2n = 2(–1/2) = –1 < n = –1/2. TRUE
III. n² = (–1/2)² = 1/4 < –n = –(–1/2) = 1/2. TRUE

Thus, I, II, and III COULD be true.

A. 7 + b/2

B. 1 + 6b

C. 1 + 3b²

D. 2

E. 3

Remember when you see word equations to get the numbers out of the words. Focus on solving for a.
2a + 8 = 6b2 + 10
2a = 6b2 + 2 ½(2a) = ½(6b2 + 2)
a = 3b2 + 1

A. 1

B. 1.5

C. 2

D. 2.5

E. 3

2x = 3x -1
Subtract 3x from each side.
-x = -1
Divide each side by -1.
x = 1

A. 5

B. 30

C. 60

D. 90

E. 120

The letter i repeats. Permutations are used.(5•4•3•2•1)/(2•1)=120

A. 2

B. 5

C. 10

D. -5

E. -2

Collecting terms in y on the left, 5y² - 10y = 0. Factorying 5y(y-2) = 0. Therefore, y = 0 and y = 2

A. 8

B. 9

C. 10

D. 11

E. 12

For the least possible value here for a + b, focus on a because it will be multiplied by 4. That gives you the chance to have a small value for b.
4a + 2b = 30
4(7) + 2(1) = 30
7 + 1 = 8

A. (-5,2)

B. (-2,5)

C. (2,5)

D. (5,2)

E. (-2,-5)

Draw yourself a sketch and think of folding the point over the x-axis.

A. 1/3

B. 1/2

C. 2/3

D. 3/4

E. 4/5

Problems like this one utilize ratios of points on lines, so we must use a least common denominator to connect values. The ratios are 1/2 and 1/3, so the least common denominator is 6. Thus we give point Z the value of 6. Y would be halfway so it would be given a value of 3. X is a third of the way, so it has a value of 2. The distance between W and Y would be 3. The distance between X and Z would be 4. Hence the ratio is 3/4.

A. -x+6y

B. -x-6y

C. x+6y

D. 13x-2y

E. 13x+10y

Removing the parentheses gives
(5x+6y)-(7x-2y)+(x-2y)
5x+6y-7x+2y+x-2y
-x+6y

A. 3

B. 4

C. 5

D. 6

E. 8

Cross multiply and solve. 15x = 5(x + 8)
15x = 5x + 40
10x = 40
x = 4

A. 10, 10, 10

B. 5, 10, 10

C. 5, 15, 15

D. 10, 15, 15

E. 5, 10, 15

Perimeter is the distance around the outside. A rectangle has opposite sides equal. One of the other sides is 5, which leaves 30 units (from the 40 in the perimeter) to be split between the other two sides.

A. -21/5

B. -5/23

C. -23/5

D. -24/5

E. -26/5

(2x/3) + 2 = (x/4)
Find a common denominator and convert 2 to fraction form.
(8x/12) + (24/12) = (3x/12)
Multiply each side by 12.
8x +24 = 3x
Subtract 8x from each side.
24 = -5x
Divide each side by -5.
x = -24/5

A. 2

B. 3.5

C. 4.2

D. 5.7

E. 6

Draw a diagram. The right triangle has a hypotenuse of 6 and the angle of 71º should be placed between the ladder and the ground. Place x on the vertical side. The correct trig. function is sin 71º = x/6.

A. 1

B. 5

C. 10

D. 11

E. 12

We want the greatest possible value for a - b we want the greatest value for "a" and the smallest value for "b." And note these are both INTEGERS.
a = 13
b = 1
13 - 1 = 12

A. 5

B. 7

C. 1

D. -5

E. -7

Remember to isolate the variable.
x + 17 = 12.
Add -17 to both sides x + 17 - 17 = 12 - 17.
Combine like terms. x = -5

A. 0

B. 1

C. -1

D. 2

E. undefined.

A vertical line has "rise" but no "run". Thus, the denominator in the rise/run formula is 0 and it is an undefined fraction.

A. x = -2

B. x = 2

C. y = -2

D. y = 2

E. y = 1

There is a formula for finding this equation without graphing the parabola. Axis of symmetry is x=-b/(2a). So we have
x=(-12)/(2)(3)=-2

A. $60

B. $72

C. $96

D. $192

E. $227

25% of the original price can be represented algebraically by (0.25)(x). This expression equals $48. So we have (0.25)(x)=48. To solve for x, divide each side of the equation by 0.25. The value for x is 192.

A. y=3x+2

B. y=x+2

C. y=3x+1

D. y=3x-1

E. y=2x-1

Use the point-slope form. y-y1=m(x-x1)
y-2=3(x-1)
y-2=3x-3
y=3x-1

A. 1/2

B. 1/3

C. 2/3

D. 4/6

E. 5/6

Numbers greater than 2 are 3,4,5 and 6.
Numbers that are even are 2,4 and 6.
Numbers in both are 4 and 6. There are 2 values out of 6.
2/6=1/3

A. 7x²-10y+10

B. x²-10y-2

C. x²+4y-2

D. -x²-4y+2

E. x²-5y-2

Line up the two expressions with 4x²-3y+4 on the top. Be sure to remember to change the signs when subtracting.

A. xz

B. zx²

C. xz²

D. xz³

E. x³z²

Given that x = y/z, we know that y = xz.
So, yz = (xz)z = xz².

A. 1

B. 2

C. 0.5

D. -1

E. -2

5c - 4 - 2c + 1 = 8c + 2
3c -3 = 8c + 2
-3 = 5c + 2
-5 = 5c
-1 = c

A. p + q - r

B. -p + q + r

C. p - q - r

D. p + q + r

E. p / q * r

Mr. Jones is presently (q + r) years old.
Therefore, he will be (p + q + r) years old p years from now.
If the problem appears confusing, choose an easy set of numbers,
e.g., p = 7; q = 50 and r = 3.
Then, Mr. Jones was 50 years old 3 years ago.
Therefore, he is 53 and will be 60 years old 7 years from now.
Note that p + q + r = 7 + 50 + 3 = 60.

A. 2

B. 3

C. -3

D. 6

E. -6

Adding the equations gives 0B + 6C = 12. Thus c = 2

A. 36/48

B. 40/64

C. 0.625

D. 15/24

E. 75/120

36/48 = 6/8 = 3/4 = 0.75, whereas 35/56 = 40/64 = 0.625 = 62.5% = 5/8.

A. 2.70%

B. 3.75%

C. 27%

D. 22.75%

E. 37.50%

Written literally: 30 = m% of 80
30 = (m/100)(80)
30 = (80m)/100
80m=3000
m=37.5

A. 2.5

B. 3.5

C. 3.6

D. 4.5

E. 5.2

0.7x + 2(x-3)=0.2x + 3
0.7x + 2x - 6 = 0.2x +3
2.7x - 6 = 0.2x + 3
2.5x - 6 = 3
2.5x = 9
x=3.6

A. 24

B. 16

C. 28

D. -16

E. -24

Be careful of this wording. "Subtracted from" means that we write -4 - 20 which equals -24.

A. 6a+6b

B. a+6b

C. -2b

D. 6b

E. 6a-6b

Remember to distribute the negative sign across the parentheses. This distribution gives 4a+2b-2a+3b-a+b. Combining similar terms gives 1a+6b = a+6b.

A. 2.78

B. 3.23

C. 3.67

D. 3.87

E. 4.21

297/3 = 99 ; 99/3 = 33 ; 33/3 = 11. So, n = 3.
Note 297/(3³) = 297/27 = 11, whereas 297/(3⁴) = 297/81 = 3.67.

A. 1

B. 3

C. 9

D. 18

E. 27

3 x 3 = 9

A. 80

B. 100

C. 1000

D. 1600

E. 2200

We substitute x² for y in the system of equations.
x³ = 40x²
x = 40
y = x² = 40² = 1600

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is C

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is D

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is D

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is C

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is D

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is C

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is C

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is D

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is B

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is D

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is E

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is C

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is B

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is B

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is A

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is C

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is B

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is A

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is C

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is A

A. 1 cube

B. 2 cubes

C. 3 cubes

D. 4 cubes

The Correct Answer is C

A.

B.

C.

D.

E.

The correct answer is C.
The paper is two folds thick where the punch occurs and therefore has two hole punches.

A.

B.

C.

D.

E.

The correct answer is D.
The paper is four folds thick where the punch occurs and therefore has four hole punches.

A.

B.

C.

D.

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The correct answer is E.
The paper is four folds thick where the punch occurs and therefore has four hole punches.

A.

B.

C.

D.

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The correct answer is C.
The paper is four folds thick where the punch occurs and therefore has four hole punches.

A.

B.

C.

D.

E.

The correct answer is B.
The paper is two folds thick where the punch occurs and therefore has two hole punches.

A.

B.

C.

D.

E.

The correct answer is D.
The paper is eight folds thick where the punch occurs and therefore has eight hole punches.

A.

B.

C.

D.

E.

The correct answer is B.
The paper is four folds thick where the punch occurs and therefore has four hole punches.

A.

B.

C.

D.

E.

The correct answer is D.
The paper is two folds thick where the punch occurs and therefore has two hole punches.

A.

B.

C.

D.

E.

The correct answer is B.
The paper is two folds thick where the punch occurs and therefore has two hole punches.

A.

B.

C.

D.

E.

The correct answer is A.
The paper was only one fold thick at the origin of the punch and therefore has only one hole punch.

A.

B.

C.

D.

E.

The correct answer is A.
The paper is six folds thick where the punch occurs and therefore has six hole punches.

A.

B.

C.

D.

E.

The correct answer is D.
The paper is eight folds thick where the punch occurs and therefore has eight hole punches.

A.

B.

C.

D.

E.

The correct answer is A.
The paper is four folds thick where the punch occurs and therefore has four hole punches.

A.

B.

C.

D.

E.

The correct answer is A.
The paper is six folds thick where the punch occurs and therefore has six hole punches.

A.

B.

C.

D.

E.

The correct answer is D.
The paper is three folds thick where the punch occurs and therefore has three hole punches.

A. 4 - 1 - 3 - 2

B. 4 - 1 - 2 - 3

C. 3 - 1 - 2 - 4

D. 3 - 2 - 1 - 4

The correct answer is A

A. 4 - 1 - 2 - 3

B. 4 - 1 - 3 - 2

C. 3 - 4 - 2 - 1

D. 3 - 1 - 2 - 4

The correct answer is A

A. 3 - 1 - 2 - 4

B. 3 - 4 - 2 - 1

C. 3 - 1 - 4 - 2

D. 2 - 3 - 1 - 4

The correct answer is C

A. 2 - 3 - 1 - 4

B. 2 - 1 - 3 - 4

C. 2 - 1 - 4 - 3

D. 2 - 4 - 1 - 3

The correct answer is D

A. 3 - 1 - 2 - 4

B. 3 - 2 - 1 - 4

C. 3 - 4 - 1 - 2

D. 3 - 2 - 4 - 1

The correct answer is B

A. 1 - 3 - 2 - 4

B. 3 - 2 - 1 - 4

C. 3 - 4 - 1 - 2

D. 3 - 2 - 4 - 1

The correct answer is D

A. 2 - 3 - 1 - 4

B. 2 - 1 - 3 - 4

C. 2 - 1 - 4 - 3

D. 2 - 4 - 1 - 3

The correct answer is B

A. 1 - 2 - 4 - 3

B. 1 - 2 - 3 - 4

C. 2 - 3 - 4 - 1

D. 2 - 1- 3 - 4

The correct answer is A

A. 3 - 1 - 2 - 4

B. 3 - 2 - 1 - 4

C. 3 - 4 - 1 - 2

D. 3 - 2 - 4 - 1

The correct answer is B

A. 3 - 4 - 2 - 1

B. 3 - 1 - 2 - 4

C. 3 - 2 - 4 - 1

D. 3 - 2 - 1 - 4

The correct answer is B

A. 4 - 1 - 3 - 2

B. 4 - 2 - 3 - 1

C. 4 - 1 - 2 - 3

D. 3 - 2 - 1 - 4

The correct answer is B

A. 2 - 1 - 3 - 4

B. 2 - 3 - 1 - 4

C. 1 - 2 - 3 - 4

D. 1 - 4 - 3 - 2

The correct answer is D

A. 1 - 3 - 2 - 4

B. 3 - 2 - 1 - 4

C. 3 - 4 - 1 - 2

D. 3 - 2 - 4 - 1

The correct answer is B

A. 1 - 3 - 2 - 4

B. 1 - 2 - 3 - 4

C. 4 - 3 - 1 - 2

D. 4 - 2 - 1 - 3

The correct answer is A

A. 4 - 1 - 3 - 2

B. 2 - 1 - 4 - 3

C. 2 - 1 - 3 - 4

D. 2 - 3 - 1 - 4

The correct answer is C

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is C.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is D.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is C.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is D.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is D.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is C.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is A.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

The Correct Answer is B.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is A. It is taken from the top / bottom angle and is rotated 90 degrees clockwise from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is C. It is taken from the side angle. The horizontal part of the L is the correct size in comparison with the other options.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is A. It is taken from the back side of the 3-D image. You will notice that the lower trapezoid is bigger than the adjoining higher trapazoid. This wider lower trapazoid gives us the wider horizontal box of the reverse L like image of option A.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is A. It is taken from the back angle and is rotated 90 degrees clockwise and inverted from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is D. It is taken from the back of the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is C. It is taken from the side angle and is flipped just a little more than 180 degrees from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is D. It is taken from the front / back angle and is flipped vertically from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is E. It is taken from the right side angle of the 3-D image. Thus only one of the protuding bodies would be seen.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is E. It is taken from the side angle and is rotated 90 degrees counter-clockwise from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is A. It is taken from the front angle and is 90 degrees counter-clockwise from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is C. It is taken from the back angle and flipped 90 degrees clockwise compared to the 3-D image. Notice the small point coming from the middle that is only shown in option C.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is E. It is taken from the back side angle and is flipped 90 degrees counter-clockwise compared to the 3-d image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is E. It is taken from the front angle and is flipped 180 degrees from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is C. It is taken from the side angle and is flipped vertically from the 3-D image.

A. Drawing A

B. Drawing B

C. Drawing C

D. Drawing D

E. Drawing E

The correct answer is B. It is taken from the side angle and is rotated 90 degrees clockwise from the 3-D image.

A. Spasms of the internal focusing muscles within the eye.

B. Genetic factors.

C. Environmental influences.

D. The eyeball growing too short.

E. The eyeball growing too long.

As the article states in lines 20-21, “However, there is also another type of spurious myopia that is due to Excessive spasms of the internal focusing muscles within the eye.”

A. Genetic factors.

B. Lights on at night.

C. Environmental influences.

D. All of the above.

E. Both A and C.

As the article states on line 5, “There is still no complete explanation for the development of myopia.”

A. Prevent myopia from developing.

B. Curing myopia.

C. Slow down the progression of myopia.

D. Both A and B.

E. None of the above.

As the article states in lines 34-35, “The challenge is to either prevent it from developing in the first place, or to slow down its progression.”

A. Wearing glasses.

B. Undergoing surgery to flatten the cornea.

C. Wearing contact lenses.

D. All of the above.

E. None of the above.

As the article states in lines 25-26, “Although true myopia is irreversible, it can be compensated for by wearing spectacles, contact lenses or undergoing surgery to flatten the cornea.”

A. Environmental influences.

B. Genetic factors.

C. Diet.

D. Inbreeding.

E. Poor light.

As the article states in lines 8-9 “This suggests strongly that environmental influences interacting with 10 genes have played a major role.”

A. Ponds.

B. Open water.

C. Water in cement containers.

D. All of the above.

E. None of the above.

As the article states in lines 25-27, "Water sections include dealing with oil slicks on open water as well as the remediation of petroleum-polluted water found in industrial retaining ponds and other concrete containments.”

A. Using bacteria to clean petroleum polluted areas.

B. Using chemicals to clean petroleum polluted areas.

C. Using information technology to clean petroleum polluted areas.

D. All of the above.

E. None of the above.

As the article states in lines 1-3, “Bioremediation is the process of using hydrocarbon-oxidizing bacteria to break down the chemical structure of petroleum compounds into less complex, fertilizer-like substances that are not hazardous or regulate.”

A. Using video.

B. Using pictures.

C. From steps one to conclusion.

D. Both A and B.

E. Both B and C.

As the article states in lines 20-21, “In a step-by-step fashion, the illustrated web feature explains how bioremediation is used to degrade petroleum pollution from soils and water.”

A. So farmers can understand bioremediation better.

B. So scientists can understand bioremediation better.

C. So governments can understand bioremediation better.

D. So non-scientists can understand bioremediation better.

E. So farmers can understand scientists better.

As the article states in lines 14-18, “BIOREMEDIATION: A Layman's Guide to Techniques and Materials," the extensive reference resource provides practical, "hands on" information for persons who need to understand the process of bioremediation to better manage outside contractors performing such work, or who want to undertake their own bioremediation projects in accordance with currently accepted scientific standards.”

A. To clean up spilled oil.

B. To clean up the ozone layer.

C. To create cleaner farming techniques.

D. All of the above.

E. None of the above.

As the article states in lines 5-7, “Since the late 1980s when it was recognized by the U.S. EPA as a viable petroleum clean-up method, bioremediation has grown into a major segment of the American environmental remediation industry.”

A. For health.

B. For treatment.

C. For longevity.

D. For commercialization.

E. For depression.

As the article states in lines 13-14, “Vitamin B12 was isolated in 1926 as the factor that treated a feared disease, Pernicious anemia.”

A. Cobalamin.

B. Pernicious anemia.

C. Hydrochloric acid.

D. Mucoprotein enzyme.

E. Calcium.

As the article states in lines 18-20, “It was later found that a mucoprotein enzyme produced by the stomach (by the parietal cells that also make hydrochloric acid) was also needed for vitamin B12 to be absorbed into the body from the intestines.”

A. Fruit.

B. Chicken.

C. Turkey.

D. Vegetables.

E. Liver.

As the article states in line 16, “(actually liver was used as the cure; it contains high amounts of B12).”

A. Aging.

B. Stress.

C. Stomach problems.

D. All of the above.

E. None of the above.

As the article states in lines 22-23, “Aging, stress, and problems with the stomach or stomach surgery weaken the body's ability to produce the "intrinsic factor".”

A. Only vitamin with cobalt.

B. Only vitamin with calcium.

C. Only vitamin with tryptophan.

D. Only vitamin with thyroid hormones.

E. It is not really a vitamin.

As the article states in lines 2-3, “B12 is unique in that it is the only vitamin that contains an essential mineral--namely, cobalt.”

A. A scientist.

B. A pharmacist.

C. A pharmacologist.

D. Both A and B.

E. Both B and C.

As the article states in lines 5-6, “a pharmacist uses knowledge which is generated by scientists to apply to real life situations.”

A. A scientist.

B. A pharmacist.

C. A pharmacologist.

D. Both A and B.

E. Both B and C.

As the article states in lines 7-9, “The pharmacist on the other hand as a professional would be the one who would work in a poison control center.”

A. A degree.

B. Judgment calls.

C. Being a scientist.

D. Both A and C.

E. Both B and C.

As the article states in lines 11-12, “It's the judgment calls based on scientific information from a wide variety of fields of study which is part of what makes a pharmacist a professional.”

A. To teach future scientists.

B. To teach future pharmacists.

C. To teach future pharmacologists.

D. Both A and B.

E. Both A and C.

As the article states in lines 21-23, “Pharmacy schools thus tend to have 2 major divisions. Pharmacy Practice, where the graduates are pharmacists, and the pharmacological or pharmaceutical sciences, where the graduates are trained as scientists.”

A. Pharmacy is a place, whereas pharmacology is a science.

B. Pharmacy is a science, whereas pharmacology is a profession.

C. Pharmacy is a profession, whereas pharmacology is a science.

D. All of the above.

E. None of the above.

While the article states in line 1, “Pharmacy is a profession, whereas pharmacology is a science.”

A. Obesity.

B. The levels of several pro-inflammatory mediators.

C. Reactive oxygen species generation by binuclear cells.

D. Reactive oxygen species generation by mononuclear cells.

E. Depression.

As the article states in lines 17-18, “insulin reduced reactive oxygen species generation by mononuclear cells.”

A. Diabetics.

B. Non-diabetics.

C. Heavier people.

D. All of the above.

E. Both B and C.

As the article states in lines 10-11, “10 obese, non-diabetic subjects.”

A. Placebos

B. Insulin Intravenous solutions

C. Non-Insulin Intravenous solutions

D. Both A and B

E. Both B and C

As the article states on line 10, “insulin and non-insulin-containing intravenous solutions.”

A. Insulin may increase the risk of heart attacks

B. Insulin may decrease the risk of heart attacks

C. Insulin may increase the risk of atherosclerosis

D. Insulin may decrease the risk of atherosclerosis

E. Insulin may increase depression

As the article states on lines 25-27, “At the very least, these results should indicate to physicians that they should not be reluctant to prescribe insulin when it is indicated for fear that it may increase the risk of heart attack.”

A. Insulin is for diabetics only

B. Insulin can work as an inflammatory

C. Insulin may increase the risk of heart attacks

D. Insulin may reduce the risk of atherosclerosis

E. Insulin may function as an antidepressant

As the article states on lines 15-19, “The investigators found that insulin reduced concentrations of the pro-inflammatory compound nuclear factor kappa-B (NF kappa-B) and increased concentrations of the compound that inhibits NF kappa-B. In addition, insulin reduced reactive oxygen species generation by mononuclear cells and produced a reduction in the levels of several other pro-inflammatory mediators.”

A. Clothing

B. Latitude

C. Longitude

D. Ozone pollution

E. Shade from buildings

As the article states on lines 7-10, “Latitude, season, and time of day as well as ozone pollution in the atmosphere influence the number of solar ultraviolet B photons that reach the earth's surface, and thereby, alter the cutaneous production of cholecalciferol.” When they are referring to Boston they are talking about winter meaning latitude.

A. Vitamin D deficiency

B. Vitamin D insufficiency

C. None of the above

D. All of the above

E. None of the above

As the article states on lines 14-15, “It is now recognized that vitamin D insufficiency and vitamin D deficiency are common in elderly people.”

A. Aging

B. Sunscreen

C. An increase in skin pigmentation

D. All of the above

E. None of the above

As the article states on lines 6-7, “An increase in skin pigmentation, aging, and the topical application of a sunscreen diminishes the cutaneous production of cholecalciferol.”

A. Osteomalacia

B. Cholecalciferol

C. Pre-cholecalciferol

D. 7-dehydrocholesterol

E. Durmalcia

As the article states on lines 4-6, “pre-cholecalciferol undergoes a thermally induced rearrangement of its double bonds to form cholecalciferol.”

A. Humans

B. Animals

C. Vertebrates

D. Invertebrates

E. Plants

As the article states on lines 1-2, “All vertebrates, including humans, obtain most of their daily vitamin D requirement from casual exposure to sunlight.”

A. Rus

B. Nestor

C. Rimbert

D. Erimbert

E. Ipaty

As the article states on lines 25-26, “What happened in the early days of the Rus" we can use Rimbert as a Prime source but the Nestor's Chronicle only as a Secondary source.”

A. Russian history is well explained by Nestor.

B. Few books have survived since the 9th century.

C. There is little written about events in the 9th century.

D. His work is corroborated with other history texts of the time.

E. His work was written in an understandable format that the common person would comprehend.

As the article states on lines 15-16, “In the later case the information given certainly is secondary material and thus from a point where we would like to have more knowledge about the 9th Century events.”

A. They have no relationship.

B. They were all written by Nestor.

C. None were written by Nestor.

D. They were written at the same time.

E. Young people do not recognize when they are depressed.

As the article states on lines 9-11, “The Nestor's Chronicle, as goes for the Ipaty Annals and the Lavrenty Annals which scholars link together with the Nestor's work, can be a transcribe good version of the sources Nestor had at hand, but that we can't be sure about that.” In other words they share sources of information from the same time period.

A. Primary history.

B. Secondary history.

C. Neither primary nor secondary histories.

D. Both A and B.

E. None of the above.

As the article states on lines 25-26, “we can use Rimbert as a Prime source.”

A. Primary history.

B. Secondary history.

C. Neither primary nor secondary histories.

D. Both A and B.

E. None of the above.

As the article states on lines 17-18, “In today's Russia the Nestor's Chronicle is a Secondary source and shall be treated methodological as such.”

A. Depression needs to be treated.

B. Depression in young people exists.

C. Depression in young people is different from depression with adults.

D. There are multiple symptoms when diagnosing depression in young people.

E. Young people do not recognize when they are depressed.

The article states discusses that young people suffer from depression.

A. Doctors do not comprehend how young people become depressed.

B. Children do not show the same signs of depression as adults.

C. Children do not know how to tell someone that they are depressed.

D. Depression in young people is often a grouping of behavioral concerns.

E. Children do not recognize that they are depressed.

As the article states on lines 8-10, “depression in young people seldom presents as a solitary problem but is commonly part of a complex pattern of behavioral concerns, which can be challenging both for diagnosis and treatment.”

A. Anxiety

B. Substance abuse

C. Bipolar mood disorder

D. Disruptive behavior disorders

E. Bulimia

As the article states on lines 11-13, “43% of depressed youth had at least 1 other concurrent diagnosis, most commonly anxiety (18%), substance abuse (14%), and disruptive behavior disorders (8%)”

A. Over eating

B. Sleeping problems

C. Separation anxiety

D. Rages of anger

E. Both B and C

As the article states on lines 4-7, “Prominent problems in youth include somatic complaints and separation anxiety,”

A. There must be proper treatment.

B. Depression must be clearly recognized early.

C. Depression is a growing concern among young people.

D. Depression is different between young people and adults.

E. Depression in young people is caused by adults.

As the article states on lines 1-2, “It is essential that depression be recognized early and accurately by primary care providers.”

A. Glaucoma is an awful disease.

B. Psychophysics is not a real science.

C. Some diseases need to have both subjective and objective analysis.

D. A doctor should be able to analyze a symptom better than a patient could.

E. A patient should be able to analyze a symptom better than a doctor could.

This article states that often medical diagnosis is a two way road needing both the patient’s subjective reports as well as the doctor’s objective analysis.

A. It is a curable disease.

B. It is a made up disease.

C. It is not a completely understood disease.

D. It is a socially acceptable disease.

E. It is defined by the patient’s symptoms as much as the patient’s description of how they feel.

As the article states on line 20, “Glaucoma thus constitutes a psychophysically consistent entity.” Psychophysics is the main concept behind this article.

A. The study of physical conditions.

B. The study of patient doctor relationships.

C. The connection between nerve action and consciousness.

D. The connection between the real world and the imaginary.

E. The connection between a person and their psychosis

As the article states on lines 6-8, “describing relationships between either normal or abnormal subject/customer/patient-subjective and objective data related to sensory functions of human beings'.”

A. Victim

B. Subject

C. Customer

D. Consumer

E. Recipient

As the article states on line 3, “customer (otherwise known by the dirty word 'patient'.”

A. A cure for glaucoma.

B. The effect of glaucoma.

C. The need for glasses in order to see properly.

D. An area of diminished vision within the visual field.

E. The need for contacts in order to see properly.

As the article states on lines 13-18, “subjectively space wise-envisioned, sightless areas…”.

A. That brand name drugs are better than generic drugs.

B. That generic drugs are better than brand name drugs.

C. That brand name drugs are different from generic drugs.

D. That there are only small differences between generic and brand name drugs.

E. That generic drugs will have a greater chance for side effects.

The article states while generic drugs and name brand drugs are similar in many ways there is still a small difference between the two.

A. How generic drugs cost less than name brand drugs.

B. How name brand drugs cost more than generic drugs.

C. How quickly the drug enters the blood after ingestion.

D. How much of the drug enters the blood after ingestion.

E. How the different active ingredients affect each other.

As the article states on lines 16-20, “There can also be occasional differences in an individual's response to a generic drug (compared to the brand-name drug), since formulation differences can affect factors such as how much actually enters the bloodstream after oral administration.”

A. Texas

B. Florida

C. California

D. Alaska

E. None

The article refers to each state but no state in particular.

A. Nausea

B. Hair loss

C. Weight gain

D. Loss of appetite

E. Depression

As the article states on line 16, “there can be occasional differences in individual side-effects (nausea, for example).”

A. Name

B. Cost

C. Active ingredients

D. Coating

E. Both A and B

As the article states in lines 10-11, “The names are different, and the price of the generic drug is usually lower than that of a name-brand drug.”

A. ₁H² + ₁H² -> ₂He⁴

B. ₀n¹ + ₁₃Al²⁷ - > ₁₁Na²⁴ + ₂He⁴

C. ₁₃Al²⁷ + ₂He⁴ -> ₁₅P³⁰ + ₀n¹

D. ₇N¹⁴ + ₂He⁴ -> ₁H¹ + ₈O¹⁷

E. ₇Al²⁴ + ₃He³ -> ₁H¹ + ₇O¹⁵

Fusion is the process of joining two light nuclei to form a heavier one. High pressure and high temperatures are used to make this happen. Tremendous energy is released when the fusion takes place. Answers B, C,D, and E are all examples of artificial transmutation or bombardment of an element by a neutron or alpha particle. Only answer A shows two light elements combining.

A. 28.

B. 56.

C. 62.

D. 90.

E. 91

Atomic mass is found by adding protons plus neutrons. 28 + 34 = 62. Electrons have a mass of nearly 0 so they are not counted in the formula for atomic mass.

A. None.

B. 0.5 million.

C. 1 million.

D. 1.5 million.

E. 2 million.

A hybrid black-coated guinea pig has a genotype of Bb. During meiosis, 50% the resulting sperm cells will receive a B and the other 50% will receive a b.

A. It is nontoxic.

B. It is denser than air.

C. It is a colorless gas.

D. It has the odor of rotten eggs.

E. It is an acidic gas which reacts with alkali and metals like silver.

Hydrogen sulfide gas not only is denser than air, smells awful, and has no color, but also is extremely toxic. Furthermore it is an acidic gas which reacts with alkali and metals like silver. This is the reason why silver jewellery turns black when exposed to polluted atmosphere for a long duration.

A. AO x BB

B. AA x BO

C. AB x OO

D. AO x BO

E. AA x BB

To get a child with blood type O, each parent must contribute the recessive blood type allele, O, as indicated in the forth cross.

A. Loss of atoms.

B. Atomic fusion.

C. Atomic fission.

D. Both fusion and fission.

E. Electrical charge.

Atomic fusion is the process of combining atoms to form elements of higher atomic weight. Often large amounts of energy is released during this process. Meanwhile, atomic fission is the process of splitting an atomic nucleus. This results in the release of large amounts of energy as well.

A. 2.81 g

B. 14.1 g

C. 22.4 g

D. 34.2 g

E. 63.3 g

Remember at STP a mole of gas occupies 22.4 liters. Density = mass / volume
or M = D x V
Substitute: 0.63 g for mass and multiply by 22.4 L to find the answer.
0.63 g x 22.4 L = 14.1 g

A. 3

B. 5

C. 7

D. 10

E. 14

On the pH scale, a pH of less than 7 is acidic; a pH of more than 7 is basic. A neutral solution such as pure water has a pH of 7.

A. Water.

B. Feces.

C. Uric acid.

D. Carbon dioxide.

E. Bile.

The four metabolic waste products are water, carbon dioxide, salts, and nitrogenous wastes. Water is necessary for many chemical reactions including hydrolysis but it is a waste product of dehydration synthesis occurring in the cells.

A. liver.

B. bone.

C. brain.

D. skeletal muscle.

E. stomach.

Under basal conditions, the liver receives about 35% of the total cardiac output, whereas the brain and skeletal muscle receive about 15%. The bone is relatively nonvascularized and receives on 5% of the total blood flow.

A. genus.

B. species.

C. community.

D. population.

E. order.

A community is all of the living organisms in a particular area at a particular time. A population is all of one species in one area at one time. A species is a group of organisms that look alike and can breed among themselves and a genus is a broader division of classification.

A. Mass only.

B. Charge, only.

C. Both charge and mass.

D. Neither charge and mass.

E. Not applicable.

Matter can never be created or destroyed but it can change its form. Mass is always conserved in chemical reactions. Because the loss of electrons (by oxidation) must equal the gain of electrons (by reduction), charge must also be conserved. Only choice C includes conservation of charge and mass.

A. increase.

B. decreases.

C. remains the same.

D. none of the above.

E. Not applicable.

Entropy is a measure of the disorder or randomness (lack of order) of a system. Molecules in gases have no definite volume and are more free to move than molecules in solids. As molecules change phase from a gas to a solid, the molecules are becoming more orderly and therefore have less disorder, less entropy. Compare the room of a teenage to the room of a parent. The teen has a more disorderly room (higher entropy). As the teen gets older and becomes a parent, the disorder in the room decreases.

A. Water is released.

B. Oxygen is released.

C. Nitrogen is absorbed.

D. Phosphorus is absorbed.

E. Phosphorus is released.

Dehydration synthesis is the process of building up molecules by removing water. A hydroxyl molecule (-OH) is broken off one organic molecule and a hydrogen atom (H) is removed from the other organic molecule. These combine to form water (H2O), a waste molecule. The two organic molecules are now unstable and bond to each other.

A. Butane.

B. Butene.

C. Benzene.

D. Butyne.

E. Bauxite.

Benzene has the formula C₆H₆ and is ring-shaped or shaped like a closed chain.

A. Thyroxine.

B. Estradiol.

C. Testosterone.

D. Prostaglandins.

E. Estrogen.

Thyroxine is an iodinated derivative of the amino acid tyrosine. Other hormones derived from amino acids include melatonin and epinephrine.

A. equal distribution of auxin in the stem.

B. unequal distribution of auxin in the stem.

C. equal distribution of a neurotransmitter in the stem.

D. unequal distribution of a neurotransmitter in the stem.

E. force of gravity.

The plant hormone, or auxin, is distributed on the side away from the light causing cells on that side to grow faster, which in turn, causes the plant to bend toward the light. If the distribution were equal, the plant would grow just upwards. Neurotransmitters are chemicals secreted by multicellular animals and are used in transmitting impulses in the nervous system.

A. iron.

B. copper.

C. Vitamin A.

D. Vitamin B.

E. Vitamin K.

Vitamin A (also known as retinol) is converted in the retina to retinal. Retinal is a light-sensitive pigment, necessary for night vision.

A. It is lighter than air.

B. It is soluble in water.

C. It has a pungent smell.

D. It reacts with ammonia forming ammonium chloride.

E. It has a boiling point of 188 K.

Hydrogen chloride gas is extremely soluble in water, has a pungent smell, has a boiling point of 188 K, and reacts with ammonia to form white fumes of ammonium chloride. It is not lighter than air. The density of the gas is about 25% heavier than air.

A. square.

B. pyramid.

C. rectangle.

D. tetrahedron.

E. circle.

Organic compounds are three dimensional and CH₄ has 4 equivalent single bonds. Tetra means 4.

A. 100 mV.

B. 50 mV.

C. 10 mV.

D. "-50 mV."

E. "-100 mV."

The normal resting potential of -50 MC is established by the sodium-potassium pump of the cell. This results in a high extracellular sodium concentration (~142 mEq / liter) and a high intracellular potassium concentration (A~141 mE~/liter).

A. Leukocytes contain hemoglobin.

B. Leukocytes are white blood cells.

C. Leukocytes move actively by amoeboid movements.

D. There are the same number of leukocytes as erythrocytes in plasma.

E. All of the above are true.

Only red blood cells carry hemoglobin, the substance responsible for oxygen and carbon dioxide transport.

A. All the time.

B. Only in the daytime.

C. Only during digestion.

D. Only in total darkness.

E. Never.

Respiration breaks down glucose, releasing the energy found within it. That energy is needed all the time for any living organism to perform life functions.

A. Ca

B. Ba

C. Au

D. Al

E. Cl

Remember the Gold Rush? Gold is found as dust or nuggets uncombined in nature.

A. how plants reproduce asexually.

B. the age of a plant.

C. which plants might have a common ancestor.

D. how plants pass favorable traits to their offspring.

E. why some plants produce hemoglobin.

Comparative biochemistry is the field of biology that deals with comparing similarities among different species DNA and proteins produced from the DNA. The more similar two different species DNA is, the closer the evolutionary link, and the more recent the two species shared a common ancestor.

A. contain identical DNA.

B. contain identical RNA.

C. are evolving into the same species.

D. have the same number of mutations.

E. may have similar evolutionary histories.

Comparative biochemistry shows that the more similar two species DNA is, the more closely related they are, and the more recently they evolved separately. This is evidenced in the fact that humans and apes have more than 99% similar DNA sequences!

A. 0.035 joules

B. 0.35 joules

C. 3,500 joules

D. 12,500 joules

E. 35,000 joules

Kilo means 10³ or 1000.
35 x 1000 = 35,000 joules.

A. A

B. B

C. D

D. E

E. K

Vitamins A,D,E, and K are fat-soluble vitamins. All the B vitamins, as well as Vitamin C, are water-soluble vitamins.

A. Water and salts.

B. Proteins and water.

C. Nucleic acids and proteins.

D. Nucleic acids and minerals.

E. Proteins and minerals.

Organic compounds contain carbon and hydrogen bonds. Both nucleic acids and proteins contain these. Water (H2O) and salts (ex: NaCl) do not.

A. have no regular arrangement.

B. are in constant straight-line motion.

C. take the shape of the container they occupy.

D. have stronger forces of attraction between them.

E. have weaker forces of attraction between them.

Liquids and gases take the shape of the container they occupy and both have no regular arrangement (remember solids have a regular arrangement). Both liquids and gases have particles that are in constant straight-line motion. But liquids have a constant volume because the forces of attraction between the particles keep them together. Gases do not have a constant volume and the forces of attraction between particles are weaker.

A. Iodine.

B. Bromine.

C. Chlorine.

D. Turpentine.

E. Fluorine.

Fluorine, chlorine, bromine, iodine, and astatine make up the halogen elements. They comprise group VIIA of the periodic table. Turpentine does not belong to the halogen family.

A. 75% normal males, 25% carrier females.

B. 50% colorblind males, 50% colorblind females.

C. 25% normal male, 25% colorblind males, 25% normal females, 25% carrier females.

D. 25% normal males, 25% colorblind males, 25% carrier females, 25% colorblind females.

E. 45% normal males, 5% colorblind males, 45% carrier females, 5% colorblind females.

A female carrier has one normal X chromosome and one defective X chromosome with respect to color blindness and a normal male has a normal X and a Y chromosome.

A. Optically inactive reactants yields optically active products.

B. Optically inactive reactants yield optically inactive products.

C. Optically active products are the result of optically inactive reactants.

D. Optically active products are the result of optically active reactants.

E. The preparation of dissymmetric compounds from symmetric reactants yields no racemic modification.

Organic molecules are such that reactants must be optically active in order for the products to be optically active. The converse, that is, the relation between inactive reactants and inactive products, is also correct. However, the products of optically active reactants may not show such activity, and this is due to racemic modification of the isomers.

A. atomic masses.

B. atomic numbers.

C. number of valence electrons.

D. number of principal energy levels.

E. All of the above.

Most of the time elements in the same group have the same number of valence electrons. Valence electrons are involved in bonding and this effects how elements react and determines their chemical properties.

A. T and S.

B. R and S.

C. Log and Sin.

D. Gamma and beta.

E. Alpha and delta.

R and S derive from the Latin words rectus, meaning right, and sinister meaning left.

A. the production of identical twins.

B. increased variability in offspring.

C. the formation of polyploid offspring.

D. nondisjunction of homologous chromosomes.

E. mutation of the genes.

The concept described in the question is crossing-over. It is one way that increases variety in offspring. It occurs during meiosis when the replicated homologous chromosomes are in close contact with each other during synapses.

A. Ammonia.

B. Aluminum.

C. Nitric acid.

D. Sulfuric acid.

E. Muriatic acid.

In the Haber process, ammonia is produced. Sulfuric acid is produced by the contact process.

A. Neon.

B. Fluorine.

C. Chlorine.

D. Nitrogen.

E. Oxygen.

The noble gases, group 8A, are the least reactive elements. Look up Neon on the periodic table and notice that it has a completely filled energy shell. When an energy shell is completely filled, the element does not form bonds with other atoms. It is inert or nonreactive. Thus Neon is a monatomic gas. Cl₂, F₂, and N₂ are diatomic gases (bond with another atom. Di means two).

A. lose protons.

B. gain protons.

C. gain electrons.

D. lose electrons.

E. remain constant.

Eliminate choices 3 and 4 because only electrons are lost or gained in oxidation-reduction reactions. Write the half reaction for Cl₂(g):
Cl₂(g) + 2e- -> 2Cl^-(s)
(In 2LiCl(s), Li has an oxidation number of +1. 2 times +1 = +2; therefore Cl must have an oxidation number of -2). Cl on the left side of the equation had an oxidation number of 0. It has to gain 2 electrons to have an oxidation of -2.

A. Tubular secretion.

B. Tubular sublimation.

C. Tubular reabsorption.

D. Active transport.

E. Glomerular filtration.

The process by which the kidney removes most of the waste products from the blood by pure filtration through the glomerular membrane is known as glomerular filtration. Hydrogen ions undergo tubular secretion into the glomerular filtrate, whereas substances such as glucose, amino acids, and potassium ions are actively reabsorbed by the tubular membrane. Sublimation is a physical change from the solid directly to the gaseous state, and does not apply to the kidney or any other organ.

A. Potassium chloride.

B. Ammonium chloride.

C. Potassium chlorate.

D. Ammonium chlorate.

E. Ammonium fluoride.

Only two different kinds of atoms make up a binary compound. Potassium chloride has K atoms and Cl atoms. All other choices have 3 kinds of atoms.

A. C(s) + O₂(g) <-> CO₂(g)

B. CaCO₃(s) <-> CaO(s) + CO₂(g)

C. 2Mg(s) + O₂(g) <-> 2MgO(s)

D. 2H₂(g) + O₂(g) <-> 2H₂O(g)

E. 2H₃(g) + O₂(g) <-> 2H₂O(g)

A change in pressure affects equilibrium when gases are involved. An increase in pressure favors the formation of the smaller number of moles of gas (shifts to the side with the smaller number of moles of gas).
There is no effect on Choice A because each side has only 1 mole of gas. Choice C has 1 mole of gas on the left, none on the right so increased pressure shifts it to the right.
Choice D and E have 3 moles of gas (2H and 1 O) on the left, 2 moles on the right, so increased pressure shifts it to the right. Choice B has no gas moles on the left and 1 mole of gas on the right, so increased pressure shifts equilibrium to the left.

A. Chemical cell.

B. Galvanic cell.

C. Electrolytic cell.

D. Electrochemical cell.

E. None of the above.

Galvanic cell, chemical cell and electrochemical cell describe the same thing: a cell that generates electricity spontaneously. Examples include flashlight and car batteries. Electrolytic cells are connected to external sources of electrical energy and are used for redox reactions that do not occur spontaneously.

A. 1

B. 2

C. 7

D. 10

E. 14

By definition pH is the negative log (logarithm) of the hydronium ion concentration. A 0.01 molar solution has a H⁺ concentration of 10^-2 (move the decimal point 2 places to the right). The value of the negative exponent (^-2) gives a pH of 2.

A. centriole.

B. centromere.

C. centrosome.

D. cell plate.

E. vacuole.

The centromere appears as a spot that holds the doubled chromosomes together and attached them to the spindle. The centrioles are the cell organelles, found in animal cells, that direct the movement of the chromosomes towards the poles. The cell plate forms in plant cells with rigid cell walls which can't pinch in to separate the two newly formed cells.Vacuoles are large membrane-bound compartments within some eukaryotic cells and can serve different purposes, such as capturing food materials or unwanted structural debris surrounding the cell,etc,

A. 5.

B. 6.

C. 8.

D. 9.

E. 11.

The number of protons is the same as the atomic number. There are 5 protons so the atomic number is also 5.

A. Vacuole.

B. Nucleus.

C. Cell wall.

D. Centriole.

E. All of the above.

Animal cells contain centrioles, cylindrical organelles that direct the movement of the chromosomes during mitosis. Plant cells tend to have bigger vacuoles and they also have rigid cell walls. Both have nuclei to control the cell.

A. Reduction involves the gain of electrons.

B. Oxidation is associated with the loss of electrons.

C. The oxidized particle (atom) shows a decrease in valence number.

D. Oxidation-reduction reactions involve the loss and gain of electrons.

E. All of the above are CORRECT.

A particle loses electrons when it is oxidized. Thus oxidation is associated with electron loss. Meanwhile, electron gain is associated with reduction. Since oxidation and reduction processes occur simultaneously, loss and or gain of electrons is to be expected. An oxidized particle therefore loses its (valence) electrons. This results in an increase in valence number.

A. Neon.

B. Sulfur.

C. Sodium.

D. Nitrogen.

E. Argon.

Elements that lose electrons form positive ions that are smaller than atoms of the same element. Therefore the ionic radius of these ions is smaller than its atomic radius. Only choice c, sodium, is a metal that will lose electrons. Neon and argon are stable and will neither lose nor gain electrons. Nitrogen and sulfur will tend to gain electrons, each having a larger ionic radius than atomic radius.

A. pesticides.

B. herbicides.

C. sex hormones.

D. wildlife refuges.

E. All of the above are biological controls.

Attracting insects with their own sex hormones, or pheromones, and then trapping them is an example of a biological control because you use only a natural mechanism of attracting them. Pesticides are chemicals that kill insects and are not normally found in nature. Herbicides are chemicals that kill plants.

A. Organic compounds can exist as isomers.

B. Organic compounds are usually soluble in organic solvents.

C. Organic compounds decompose at relatively lower temperatures than inorganic compounds.

D. Reactions involving organic compounds always proceed faster than those involving inorganic compounds.

E. All of the above are correct.

It is important to understand the chemical and physical properties of organic and inorganic compounds as well as their chemical behaviors. Organic compounds exist as isomers. They decompose when heating at lower temperatures. They are also soluble in other organic solvents. Organic compounds, however, do not react with each other faster than inorganic compounds.

A. 1 X 10^-8 m.

B. 1 X 10^-9 m.

C. 1 X 10^-10 m.

D. 1 X 10^-11 m.

E. 1 X 10^-12 m.

The angstrom (A⁰ is 1 X 10^-10 m., or 1 X 10^-8 cm., the average radius of an atom.

A. Urea.

B. Glucose.

C. Estrogen.

D. Creatinine.

E. All of the above.

Urea and creatinine normally pass through the kidney in the glomerular filtrate. Meanwhile, normally all the glucose in the glomerular filtrate is actively reabsorbed by the kidney in the normal individual.

A. producing chlorophyll for plants.

B. adding carbon dioxide to the environment.

C. adding oxygen to the environment.

D. storing energy in the bonds of inorganic compounds.

E. manufacturing food from carbon dioxide and oxygen.

Heterotrophs developed the ability to break down glucose, releasing the energy found within. In this process, carbon dioxide is released as a waste product. Autotrophs were able to use this waste gas in photosynthesis.

A. 1.

B. 2.

C. 3.

D. 4.

E. 5.

736 kJ/mol is the value for Mg. Use the Periodic Table to find Mg. It has 2 valence electrons.

A. fuels.

B. shields.

C. coolants.

D. moderators.

E. inhibitors.

Do not be tricked into picking water as a coolant. The question did not ask for a coolant; it asked for "traffic cops" that slow down or moderate or keep "the lid on" a reaction.

A. Saturated fatty acids contain one or more double carbon bonds.

B. Most fatty acids in nature have an even number of carbon atoms.

C. Most fatty acids in nature have an odd number of carbon atoms.

D. Most fats containing unsaturated fatty acids are solids at room temperature, whereas fats containing saturated fatty acids are liquids.

E. Fats yield approximately 50% as much energy as do carbohydrates in humans.

Saturated fats have all their carbon bonds saturated with hydrogen. Because of this, there are no double bonds. As the number of double bonds increases, the melting point of the fat decrease. Therefore, saturated fats are normally solids at room temperature, whereas unsaturated fats are liquids. Complete oxidation of 1 g. of fat yields about 9 calories, whereas 1 g. of carbohydrate yields about 4 calories.

A. Boiling point.

B. Vapor pressure.

C. Freezing point.

D. Specific gravity.

E. Osmotic pressure.

Colligative properties of solutions is determined by the amount and concentration of the solute(s) in the solution. Therefore the boiling point, vapor pressure, osmotic pressure, and freezing point are all colligative properties. Specific gravity compares the weights of substances to that of water, is not a colligative property.

A. phylum.

B. species.

C. community.

D. population.

E. domain.

A community is all of the living, or biotic, things in one place at one time. A population is just one species in one area at one time. A phylum is a subdivision of a kingdom. A domain is a sphere of activity.

A. 10

B. 63

C. 150

D. 456

E. 630

Use H = mcT (change in temp).
H = 15 g x 4.2 J/g°C x 10°C or 630 joules.

A. One cell without a nucleus.

B. One cell with two identical nuclei.

C. Two cells, each with one nucleus.

D. Two cells, each without a nucleus.

E. Three cells, each without a nucleus.

Mitosis is the division of the nucleus into two nuclei with identical sets of DNA. Cytokinesis, or cytoplasmic division, is when the cytoplasm and other organelles split in to, producing two new cells. If only mitosis occurred, you would get one cell with two identical nuclei.

A. 100 K

B. 200 K

C. 248 K

D. 272 K

E. 298 K

Change Celsius to Kelvin by adding 273.
25°C + 273 = 298 K .

A. Hydrogen sulfide.

B. Hydrogen sulfate.

C. Ammonium sulfide.

D. Ammonium sulfate.

E. Aluminium hydroxide.

Only two different kinds of atoms make up a binary compound. Hydrogen sulfate has 3 kinds of atoms (H, S, O). Ammonium sulfate has 3 kinds (N, H, S). Aluminium hydroxide has 3 kinds (Al, O, H). Ammonium sulfate has 4 kinds (N, H, S, O).

A. can cause genetic changes to occur.

B. is an agressive defense mechanism.

C. prevents germination within the seed pod.

D. attracts insects that aid in pollination.

E. aids in the dispersal of the species.

By forcefully ejecting their seeds, it ensures that they get scattered all over, and are hopefully carried by the wind to other appropriate locations to germinate.

A. careful microscopic examinations of genes and chromosomes.

B. breeding experiments with many generations of fruit flies.

C. analysis of the offspring produced from many pea plant crosses.

D. dissections to determine how fertilization occurs in pea plants.

E. travels to the Galapagos islands.

Gregor Mendel used simple mathematics and kept careful records of his pea plant crossed to arrive at his conclusions. He did know about genes and did no dissections.

A. The I ^- ion is oxidized, and its oxidation number increases.

B. The I ^- ion is oxidized, and its oxidation number decreases.

C. The I ^- ion is reduced, and its oxidation number increases.

D. The I ^- ion is reduced, and its oxidation number decreases.

E. The I ^- ion and its oxidation number remain constant.

First, determine what happens to the I ^- ion. On the left side of the equation, the I ion is negatively charged. On the right side of the equation, the I ion is neutral. The I ion lost its negative charge by losing an electron. By definition oxidation occurs when electrons are lost .The ion becomes more positive (has a higher oxidation state). Pretend electrons are dollars. If you lose a dollar, you have less money. If you find a dollar or are given one, you no longer have a negative cash flow. You can also look at the half reaction of the I ^- ion:
2I ^-(aq) -> I₂(s) + 2 e^-:
each I ^- ion loses an electron, increasing its oxidation number from -1 to 0.

A. sodium.

B. calcium.

C. potassium.

D. magnesium.

E. lithium.

Potassium is the most abundant electrolyte inside the cell (intracellular fluid).

A. methane.

B. ammonia.

C. methanol.

D. aluminum.

E. benzene.

Elements cannot be decomposed into simpler substances because an element by definition is one kind of substance, one kind of atom. The other choices are compounds--composed of two or more elements.

A. is derived from photosynthesis.

B. regulates breathing rate by its effect on the medulla.

C. causes inflammation of the tissues of the bronchial tubes.

D. regulates gastric acid production by forming carbonic acid.

E. causes increased production of hydrochloric acid.

The medulla oblongata at the base of the brain increases breathing rate if the amount of carbon dioxide increases. It does not regulate breathing rate by checking oxygen content.

A. mass number.

B. atomic number.

C. number of protons.

D. number of electrons.

E. number of neutrons.

Strontium has 2 valence electrons in its outer shell. It is easier to lose these 2 electrons than it is to gain 6 more electrons. Losing 2 electrons will give the strontium ion a charge of +2. Ions are charged atoms. An neutral atom has no charge because it has the same number of protons and electrons. An ion can have more or less electrons than an atom of the same element does. A strontium atom would have 2 more electrons than a strontium ion.

A. an allele.

B. RNA.

C. a phenotype.

D. a chromosome.

E. a genotype.

A genotype is the actual genetic makeup that is expresses in a phenotype. An allele is a gene located at a specific location on corresponding homologous chromosomes.

A. 27.8 g / mL

B. 6.6 g / mL

C. 4.2 g/ mL

D. 3.4 g/ mL

E. 3.0 g / mL

D = M / V
Step 1: find the volume of the object by subtracting 21.2 mL from 27.8 mL or 6.6 mL
Step 2: D = M / V or D = 22.4 g / 6.6 mL = 3.4 g / mL

A. GG x Gg

B. GG x GG

C. Gg x gg

D. gg x gg

E. Gg x GG

If each parent contributes only one gene of its pair to the offspring, the parental genes of Gg x gg would produce the following offspring: Gg or gg. Gg would appear gray while gg would produce black fur. The first cross would produce only gray squirrels since one parent always gives a G, masking the g given by the other parent. The second cross only has G so all offspring would have to be gray, and the last cross would not produce any gray squirrels since there are no G.

A. decrease.

B. increase.

C. remain the same.

D. none of the above.

E. Not applicable.

LeChatelier's principle: if a stress (like a change in temp. or concentration) is added to a system at equilibrium, the reaction is shifted in a way that uses up the stress. Increase the left side, the right side tries to use up the extra N₂(g). But to make more NO(g), you need to use some of the O₂(g) from the left side--remember you did not add any extra O₂(g)--so the amount of O₂(g) has to decrease.

A. addition.

B. fermentation.

C. substitution.

D. esterification.

E. reduction.

Addition usually involves adding one or more atoms at a double or triple bond. Here H₂ combines with CH₂CH₂ changing ethene (double bond) into ethane (single bond).

A. heat of reaction.

B. activation energy.

C. electrical potential.

D. potential energy of the reactants.

E. potential energy of the products.

A catalyst speeds up a chemical reaction by lowering (decreasing) the activation energy.

A. Cr³⁺ + 3e^- -> Cr(s).

B. Cr³⁺+ -> Cr(s) + 3e^-.

C. Cr(s) -> Cr³⁺ + 3e^-.

D. Cr(s) + 3e^- -> Cr³⁺.

E. Cr + 3e^- -> Cr⁴⁺.

In reduction half reactions, the oxidation number is lowered because electrons are added. When Cr³⁺ gains electrons, the negative charge of the electron cancels out the positive charge and an uncharged Cr (Cr^o) atom results.

A. A-> C-> E-> B-> D

B. E-> D-> C-> A-> B

C. C-> A-> E-> D-> B

D. C-> E-> D-> A-> B

E. D-> C-> A-> B -> E

First pour water into a beaker and take the temperature of the water. Heat of solution will be measured by determining the change of temperature after the solute (NaOH) is added to the water (and stirred). Choices 2 and 4 do not make sense as the solute NaOH is added before the temperature of the water is taken: no change of temperature can be measured this way. Choice 1 takes the temperature of the water before it is added to the beaker. Where does the water come from? Then the final temperature is taken before the solution is stirred. This method gives inaccurate readings as the thermometer could be reading a "hot" or "cold" spot in the unstirred liquid.

A. decrease.

B. increase.

C. remain the same.

D. none of the above.

E. Not applicable.

Half-life depends on time. Factors like temperature, pressure, and volume have no effect.

A. Iodine and bromine.

B. Iodine and fluorine.

C. Chlorine and bromine.

D. Chlorine and fluorine.

E. Chlorine and iodine.

Iodine is a solid and bromine is a liquid. The gaseous halogens are chlorine and fluorine.

A. This change increases the concentration of the reactant.

B. This change increases the density of the reactant particles.

C. This change increases the mass of the reactant particles.

D. This change alters the electrical conductivity of the reactant particles.

E. This change exposes more reactant particles to a possible collision.

Reaction rate is affected by nature and concentration of reactants, temperature, surface area and a catalyst. Increasing surface area exposes more particles to contact with reactants, increasing the number of particle collisions. Imagine a cube of chocolate 3 feet by 4 feet and 2 inches thick. Take the same block and make it into a 24 candy bars. A class of chemistry students could eat the bars faster than the single chocolate block.

A. synthesize its food from inorganic compounds.

B. feed upon its group species.

C. feed upon carbohydrates produced by autotrophs.

D. feed upon available nutrients in the environment.

E. carry on photosynthesis instead of respiration.

After the aggregates became so large, some developed the ability to ingest smaller organic molecules. This is heterotrophic nutrition. This set up competition as the seas became filled, so some developed the ability to change light energy into usable energy called glucose. This is autotrophic nutrition and according to the heterotroph hypothesis, autotrophic nutrition came after heterotrophic nutrition.

A. Sodium influx into the cell.

B. Sodium efflux out of the cell.

C. Potassium influx into the cell.

D. Potassium efflux out of the cell.

E. Potassium influx out of the cell.

A large sodium (NA+) influx into the cell will result in the resting potential (-50 mV) becoming more positive and thereby initiates the action potential. Potassium is already in very high concentration (~141 mEq/ liter) intracellularly and will not enter the cell.

A. requires energy sources such as ATP.

B. can be inhibited by metabolic poisons such as cyanide.

C. causes a substance to move from a lower to a higher concentration.

D. is a major process by which uncharged molecules can move through membranes.

E. causes the transport protein to change shape and mutate.

Passive diffusion of a substance requires no energy (ATP). Thus it will not be affected by a metabolic inhibitor (cyanide). As well, passive diffusion transfers uncharged (more lipid-soluble) molecules from a higher to a lower concentration.

A. an increase in the sewage content of the lake.

B. a decrease in the amount of phosphates in the lake.

C. an increase in the amount of PCB pollution in the lake.

D. a decrease in the amount of dissolved oxygen in the lake.

E. a decrease in the water level of the lake.

Discharging hot water into a lake is called thermal pollution. Besides the effect on the living organisms, warm water holds less oxygen than cool water.

A. hormones.

B. neurohumors.

C. gastric juice.

D. lachrymal fluid.

E. enzymes.

Hormones are chemical regulators that control many activities in animals, including metamorphosis of tadpoles into frogs and caterpillars into butterflies.

A. Monocytes.

B. Lymphocytes.

C. Eosinophils.

D. Neutrophils.

E. Basophils.

Lymphocytes are white blood cells that are produced by lymphoid tissue. Neutrophils, monocytes, eosinophils, and basophils are white blood cells produced by the bone marrow along with red blood cells.

A. Oxygen.

B. Hydrogen.

C. Carbon dioxide.

D. Hydrochloric acid.

E. Nitrogen.

Heterotrophs must ingest, digest, and egest preformed organic matter. Autotrophs evolved from heterotrophs by developing a way to make their own organic food by changing water and carbon dioxide into glucose.

A. 22.4 L Ar

B. 28.0 L of N₂

C. 32. 0 L of H₂

D. 36.8 L of He

E. 44.8 L of He₂

At the same temperature and pressure, equal volumes of gases have an equal (the same) number of molecules. This is known as Avogadro' hypothesis.

A. use of biocides.

B. preservation of species.

C. exploitation of species.

D. use of biological controls.

E. control over rural areas.

One positive thing humans are doing is trying to preserve species that may be endangered. Wildlife refuges (where hunting is not allowed) and game laws (that limit the number of organisms hunted) have both helped.

A. 6.89 %

B. 14.50 %

C. 26.10 %

D. 36.85%

E. 62.90 %

One mole of the hydrate given has a mass of 286. The 10 water moles are part of the hydrate. The water has a mass of 18 g x 10 or 180 g.
Percent of water = 180g (water mass) / 286 (mass of compound) x 100% = 62.9%

A. Saponification.

B. Neutralization.

C. Polymerization.

D. Oxidation-reduction.

E. Decomposition reaction.

When certain metals combine with oxygen in the air, a metallic oxide results. Iron oxide is commonly called rust. Other metals such as copper also oxidize. (The Statue of Liberty in NY Harbor was not always green). The term oxidation-reduction is used to describe the addition-removal of oxygen. Oxidation is the process by which a substance loses one or more electrons.

A. Electrolytes only.

B. Nonelectrolytes only.

C. The total number of particles present in solution.

D. Chemical transport.

E. None of the above.

Osmotic activity is a function of the total number of particles present in a given solution. These particles can be electrolytes or nonelectrolytes. Electrolytes usually will dissociate into their component ions increasing the number of particles in a solution.

A. 1:3

B. 1:4

C. 3:1

D. 4:1

E. 4:3

Four viable sperm cells are produced from one primary sex cell, whereby only one viable egg cell is produced, due to the unequal division of cytoplasm and the formation of polar bodies, which wither and die.

A. increasing need for new antibiotics.

B. increasing number of individuals in the human population.

C. decreasing number of new fossils discovered in undisturbed rock layers.

D. decreasing number of transmittable diseases.

E. decreasing activity of photosynthetic organisms due to warming of the atmosphere.

It is believed that certain pathogens have adapted to the antibodies that we have, becoming resistant. These species are actually evolving, or changing through time, to include the resistant gene for the antibiotics. Therefore the pathogens with the gene survive, and pass on the resistant gene, which causes it to increase.

A. Progesterone.

B. Vasopressin.

C. Aldosterone.

D. All of the above.

E. None of the above.

Progesterone is secreted by the corpus luteum and acts with estradiol to regulate the estrous and menstrual cycles.

A. CaCl₂₂

B. CrCl₃

C. NaOH

D. KBr

E. H₂O₃

Colored aqueous solutions are a characteristic of transition compounds. The only transition compound in the above choices is CrCl₃.

A. Methyl.

B. Methypol.

C. Methanol.

D. Methylene.

E. Methane.

Methylene (CH2) exists as a discrete molecule and is highly reactive.

A. Saturation points.

B. Melting points.

C. Boiling points.

D. Double points.

E. Triple points.

Because hydrocarbons differ in boiling points, fractional distillation separates the parts (fractions) of the mixture by heating the hydrocarbons in long columns until they form gases (vaporization). At various points in the tower the fractions (hydrocarbons) condense and are removed. This is possible because the hydrocarbons have different boiling points.

A. 7, 000.

B. 15, 813.

C. 23, 352.

D. 75, 251.

E. 158, 130.

The heat of vaporization for water is 2259 J/g.
Multiply heat of vaporization by the grams to find the calories absorbed in vaporizing water.
70 g x 2259 J/g = 158, 130 J.