MATH

Question 1

1/0.1

Answer 1

1/0.1= 1*(10/1)= 10

Question 2

-1/0.1

Answer 2

-1/0.1= -1*(10/1)= -10

Question 3

10/0.5

Answer 3

10/0.5=(10)*(2/1)=20

Question 4

(1.46x10^3)+(6.1x10^2)

Answer 4

(1.46x10^3)+(6.1x10^2)=(1.46x10^3)+(0.61x10^3)=2.07x10^3

Question 5

(-6.4x10^-6)/(-1.6x10^-19)

Answer 5

4x10^13

Question 6

5⨉10-20 J / 5⨉10-9 m

Answer 6

= 10^-11 N

Question 7

10^-11 N to pN

Answer 7

10 ⨉ 10-12 N = 10 pN There is 10^-12N/pN so… (10^-11N)/(10^-12N/pN)=10^1=10pN

Question 8

How many Hertz are in 1 MHz?

Answer 8

10^6Hz

Question 9

How many Hertz are in 1 GHz?

Answer 9

10^9Hz

Question 10

Estimate √(1/5)

Answer 10

√(1/5) about 0.45

Question 11

Estimate √(1/3)

Answer 11

0.57

Question 12

Estimate √(1/4)

Answer 12

0.5

Question 13

Estimate √(1/2)

Answer 13

0.7

Question 14

What is the total resistance supplied by three resistors in parallel with resistance values of 4 Ω, 20 Ω, and 5 Ω, respectively?

Answer 14

Total resistance for resistors in parallel is calculated according to the equation: 1/Rtotal = 1/R1 + 1/R2 + 1/R3…+ 1/Rn. Therefore, the total resistance of the circuit in question is: 1/Rtotal = 1/4 Ω + 1/20 Ω + 1/5 Ω 1/Rtotal = 5/20 Ω + 1/20 Ω + 4/20 Ω 1/Rtotal = 10/20 Ω 1/Rtotal = 1/2 Ω Rtotal = 2 Ω

Question 15

Solve: (3/4 × 2/7)/ (6/5)

Answer 15

This answer choice is correct. First, we multiply 3 x 2 and 4 x 7, resulting in 6/28. To divide this by 6/5, we multiply by the reciprocal, so (6/28) x (5/6) = 5/28. This is shown below: (3/4 × 2/7) ÷ 6/5 6/28 ÷ 6/5 6/28 × 5/6 5/28

Question 16

What is 0.32?

Answer 16

0.32 is 0.09

Question 17

If a circuit produces 20 watts of power and has a total resistance of 2 Ω, how much current is flowing through the circuit?

Answer 17

The best equation to use here is P=I2R. Since all of the given values are already in the correct SI units, we can simply plug them in, yielding: 20 W = I2(2 Ω) Dividing both sides by 2 gives: 10 = I2 Though the square root of 10 is not a value that you should have memorized, you should know that the square root of 9 is 3, so the square root of 10 should be slightly larger, or about 3.2 amperes. This corresponds to choice A.

Question 18

What is the concentration of carbonate in a saturated solution of BaCO3 at standard conditions given that the Ksp is 2.6 x 10-9?

Answer 18

The first step in any solubility question is to write out the dissociation equation. Here, this would be: BaCO3 (s) 🡪 Ba2+ (aq) + CO32- (aq). This results in a solubility constant expression of Ksp = [Ba2+][CO32-]. We know that for every “x” amount of BaCO3 that dissociates, we get “x” amount of Ba2+ and “x” amount of CO32-. We can therefore plug x into the Ksp equation for both terms, yielding 2.6 x 10-9 = [x][x], which simplifies to 2.6 x 10-9 = x2. Our final step is to take the square root of our scientific-notation term. The easiest way to do this is to try to turn the coefficient into a value whose square root can be taken easily. Here, 2.6 x 10-9 can be rewritten as 26 x 10-10. 26 is extremely close to 25, which we know has a square root of 5. Finally, we take the square root of 10-10, which is the same as (10-10)1/2. We divide -10 by 2, yielding -5. Putting this all together, our final answer is approximately 5 x 10-5M.

Question 19

The molar solubility of manganese (II) hydroxide is approximately 3.7 x 10-5. What is the Ksp of this compound?

Answer 19

Manganese (II) hydroxide has a formula of Mn(OH)2, meaning that its dissociation equation is: Mn(OH)2 (s) 🡪 Mn2+ (aq) + 2OH- (aq). The Ksp expression is thus Ksp = [Mn2+][OH-]2. Since molar solubility refers to the number of moles of solute that are dissolved in a saturated 1L solution, this is the amount of Mn(OH)2 that dissolves. We can call this “x” for the time being. For every “x” amount of Mn(OH)2 that dissolves, we get “x” amount of Mn2+ ions and “2x” amount of OH-. Plugging this into the Ksp expression yields Ksp = (x)(2x)2. Since we have to square both terms in “2x”, this results in Ksp = (x)(4x2). Since we can add exponents when we multiply, this further simplifies to Ksp = 4x3. The question stem tells us that the molar solubility of manganese (II) hydroxide is approximately 3.7 x 10-5, so we can plug this in as “x”, yielding Ksp = (4)(3.7 x 10-5)3. First, we must cube what is inside the parentheses. 3.7 can be rounded to 4, and 43, or 4 x 4 x 4, is equal to 64. Next we cube 10-5 by multiplying the exponent by 3, yielding 10-15. This now leaves us with Ksp = (4)(64 x 10-15). 64 x 4 = 256, yielding 256 x 10-15. Putting this back into scientific notation gives 2.56 x 10-13.

Question 20

Solve the following logarithm: log(10,000) = ?

Answer 20

If no base is written for a logarithm, we can assume it is base 10. This question can therefore be rewritten as log10(10,000). This can be rearranged to:10x = 10,000. 104, or 10 x 10 x10 x 10 is equal to 10,000, so 4 must be our answer.

Question 21

Solve the following logarithm: log(1) = ?

Answer 21

The log of 1 is always 0. This is something you should memorize for the MCAT.

Question 22

What is the approximate pH of a 0.003 M solution of HBr?

Answer 22

Since HBr is a strong acid, we can simply plug the concentration given into the formula pH = - log[H+]. This results in pH = - log (0.003M), which can be rewritten in scientific notation as pH = -log(3 x 10-3). 3 x 10-3 must be somewhere between 1 x 10-2 and 1x10-3. Since these exponents are 2 and 3, this means we are looking for a pH value between 2 and 3. The only answer choice that falls in this range is 2.5.

Question 23

How many times more intense is a 90 dB sound than a 60 dB sound?

Answer 23

Correct Answer The decibel equation is dB = 10log(I/I°), where I is the intensity of the sound being measured and I° is the reference intensity. The difference between 90 dB and 60 dB is 30 dB, so we can plug this into our equation: 30 dB = dB = 10log(I/I°) Dividing each side by 10 yields: 3 dB = log(I/I°) Since no base is specified, this log must be base 10, so we can rearrange this equation like this: 103 = (I/I°) 103 is equal to 1000, making D the correct answer.

Question 24

What is the concentration of silver (I) in a saturated solution of AgBr at standard conditions given that the Ksp is 5.35 x 10-13?

Answer 24

The first step in any solubility question is to write out the dissociation equation. Here, this would be: AgBr (s) 🡪 Ag+ (aq) + Br- (aq). This results in a solubility constant expression of Ksp = [Ag+][Br-]. We know that for every “x” amount of AgBr that dissociates, we get “x” amount of Ag+ and “x” amount of Br-. We can therefore plug x into the Ksp equation for both terms, yielding: 5.35 x 10-13 = [x][x] which simplifies to 5.35 x 10-13 = x2 Our final step is to take the square root of our scientific-notation term. The easiest way to do this is to try to turn the coefficient into a value whose square root can be taken easily. Here, 5.35 x 10-13 can be rewritten as 53.5 x 10-14. 53.5 is between 49, which is 72, and 64, which is 82. Since 53.5 is closer to 49, its square root should be closer to 7 than 8. We can estimate this at 7.3 or so. Finally, we take the square root of 10-14, which is the same as (10-14)1/2. We divide -14 by 2, yielding -7. Putting this all together, our final answer is approximately 7.3 x 10-7 M.

Question 25

The molar solubility of strontium iodate is approximately 3.05 x 10-3. What is the Ksp of this compound?

Answer 25

Strontium iodate has a chemical formula of Sr(IO3)2, meaning that its dissociation equation is: Sr(IO3)2 (s) 🡪 Sr2+ (aq) + 2IO3- (aq) The Ksp expression is thus Ksp = [Sr2+][IO3-]2. Since molar solubility refers to the number of moles of solute that are dissolved in a saturated 1L solution, this is the amount of Sr(IO3)2 that dissolves. We can call this “x” for the time being. For every “x” amount of Sr(IO3)2 that dissolves, we get “x” amount of Sr2+ ions and “2x” amount of IO3-. Plugging this into the Ksp expression yields: Ksp = (x)(2x)2 since we have to square both terms in “2x”, this results in Ksp = (x)(4x2). which simplifies to Ksp = 4x3 The question stem tells us that the molar solubility of strontium iodate is approximately 3.05 x 10-3, so we can plug this in as “x”, yielding: Ksp = (4)(3.05 x 10-3)3 First, we must cube what is inside the parentheses. 3.05 can be rounded to 3, and 33, or 3 x 3 x 3, is equal to 27. Next we cube 10-3 by multiplying the exponent by 3, yielding 10-9. This now leaves us with Ksp = (4)(27 x 10-9). 27 x 4 = 108, yielding 108 x10-9. Putting this back into scientific notation gives 1.08 x 10-7. This is closest to choice C.

Question 26

Solve the following logarithm: log(1/2) = ?

Answer 26

-0.3 The log of a positive number less than 1 (i.e. a decimal or a fraction) is always a negative number. Since this is the only negative answer choice, B must be the answer.

Question 27

What is the approximate pH of a 0.75 M solution of HNO3?

Answer 27

Since HNO3 is a strong acid, we can simply plug the concentration given into the formula pH = - log[H+]. This results in pH = - log (0.75M), which can be rewritten in scientific notation as pH = -log(7.5 x 10-1). 7.5 x 10-1 must be somewhere between 1 x 10-1 and 1 x 100, which is simply 1. Since these exponents are 1 and 0, this means we are looking for a pH value between 0 and 1. This narrows it down to A and B. Since 7.5 x 10-1 is very close to 10 x 10-1, or 1 x 100, we should choose the answer closer to 0, which is A.

Question 28

What is the pH of a 2M solution of acetic acid? The Ka of acetic acid is 1.8 x 10-5.

Answer 28

To solve this problem, we first need to determine the concentration of each species at equilibrium. To do this, we can use an ICE table. First, write the dissociation equation: HC2H3O2 🡪 H+ + C2H3O2- If you don’t know the formula for acetic acid, you can always use simply “HA” instead. Next, we fill in our I, or “initial,” row. Initial concentration of benzoic acid is 2 M, and initial concentrations of our product ions are zero. For our C, or “change,” row, we can say that some “x” amount of benzoic acid dissociates, and that same amount of each product ion is formed. Finally, we simply add the I and C values in each column to obtain that column’s E, or equilibrium, value. The resulting chart looks like this: HC2H3O2 🡪 H+ + C2H3O2- I 2 0 0 C -x +x +x E 2 – x x x Now the equilibrium concentrations can be plugged into the equilibrium expression for the dissociation of benzoic acid, resulting in: Ka = 1.8 x 10-5 = x2/(2 – x) For the MCAT, we can almost always assume that x is small, so we drop the “-x” term from the denominator, yielding: 1.8 x 10-5 = x2/2. Then, multiply 1.8 x 10-5 by 2 to get 3.6 x 10-5. Finally, take the square root of this number. This can be accomplished the most easily by moving the decimal so that we have 36 x 10-6. We know the square root of 36 is 6. To take the square root of the 10-6 portion, we divide the exponent by 2, resulting in a final value of 6 x 10-3. This is our x, which represents the concentration of H+ ions at equilibrium. We can then plug this into the equation pH = -log[H+]: pH = -log(6 x 10-3) Using estimation, we see that 6 x 10-3 falls somewhere between 1 x 10-3 and 1 x 10-2, so the answer should be between 3 and 2. Only choice C meets this cri

Question 29

How many times more intense is a 25 dB sound than a 5 dB sound?

Answer 29

The decibel equation is dB = 10log(I/I°), where I is the intensity of the sound being measured and I° is the reference intensity. The difference between 25 dB and 5 dB is 20 dB, so we can plug this into our equation: 20 dB = dB = 10log(I/I°) Dividing each side by 10 yields: 2 dB = log(I/I°) Since no base is specified, this log must be base 10, so we can rearrange this equation like this: 102 = (I/I°) 102 is equal to 100, making C the correct answer.

Question 30

How many μg are in 20 dg?

Answer 30

For metric conversions, it is often easiest to use scientific notation. 20 dg is 20 decigrams, which is the same as 20 x 10-1 grams. We can rewrite this in proper scientific notation as 2 x 100 grams, or simply 2 grams. 1 μg, or 1 microgram, is equal to 1 x 10-6 grams. We can thus do the following calculations: (2 x 100 g)/(1 x 10-6 g) = 2 x 106 2 x 106 is the same as 2,000,000

Question 31

Urbanium, a planet with twice the circumference and twice the mass of Earth, has an acceleration due to gravity at its surface that is closest to: 5m/s2 10m/s2 14 m/s2 20 m/s2

Answer 31

5m/s2 If Urbanium has a circumference that is two times larger than that of Earth, it must also have a radius that is twice as large. While we do not need to plug in actual values to solve this problem, we should be familiar with the equation F=Gmm/r2 Using this relationship, we see that a doubling of the planet’s mass should double the gravitational force, but a doubling of its radius should decrease it by a factor of 4. As a result, an object on Urbanium must experience a gravitational force (and thus acceleration) that is half what it would be on Earth.

Question 32

What is the energy of a photon of light with a wavelength of 430 nm? Note that Planck’s constant is 6.62 x 10-34 m2kg/s.

Answer 32

This answer choice is correct. First, we must convert 430 nm into meters, giving 430 x 10-9 m. We can then turn this into proper scientific notation, which would be 4.3 x 10-7m. Next, we can plug all of our values into E = hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength in meters. E = ((6.62 x 10-34 m2kg/s)(3.0 x 108 m/s))/(4.3 x 10-7m) We need to round these values before we can do our calculations. Let’s round 6.62 to 6.5 (though 7 would likely also be okay), 3 can stay as 3, and 4.3 becomes 4. This yields: E = ((6.5 x 10-34 m2kg/s)(3 x 108 m/s))/(4 x 10-7m) E = (19.5 x 10-26 m3kg/s2 )/(4 x 10-7m) We can then round 19.5 to 20 and divide by 4: E = 5 x 10-19 J

Question 33

A 1500-kg racecar accelerates from 0 m/s to 30 m/s in 4.2 seconds. How much more work is done by that car than by an identical one which accelerates from 10 m/s to 16 m/s in 8 seconds?

Answer 33

According to the work-energy theorem, work is equal to the change in kinetic energy throughout a process. The first racecar does not possess initial kinetic energy, so its ΔKE is equal to ½ (1500 kg)(30 m/s)2 = ½ (1.5 × 103)(9 × 102) = 6.75 × 105 J (note that rounding would absolutely be acceptable in a test-day scenario). The change in kinetic energy of the second car is equal to ½ (1500 kg)(16 m/s)2 – ½ (1500 kg) (10 m/s)2, or ½ (1500) (256 – 100). Since these numbers are even trickier to work with, convert ½ (1500) into 750 and use scientific notation. We get (7.5 × 102)(1.56 × 102), or approximately 11 × 104, or 1.1 × 105, J. Finally, the question asks for the difference in work exerted by the two cars. 6.75 × 105 J – 1.1 × 105 J = about 5.7 × 105 J.

Question 34

The pressure drop across a tube with a radius of 10 cm and a length of 2 m is 200 Pa. What is the flow rate of the water in the tube? Note that the viscosity of water is approximately 1 x 10-3 Pa∙s.

Answer 34

3.9 m3/s

Question 35

Given the equation, If the length of the tube is doubled and its radius is quadrupled, how will this affect the velocity of the fluid flowing through it?

Answer 35

The velocity of the fluid will increase by a factor of 8. ## Footnote For this question, we will need to look at equation 2. Note that this is the equation for v-1, or the inverse of velocity, so increasing the numerator will actually decrease velocity and increasing the denominator will increase velocity. We can thus rewrite a simplified version of the equation, wherein we ignore all of the other terms not being changed by designating them all as k: 1/v= kL/r2 Rearranging this to make it v instead of 1/v yields: v=kr2/L We can now see that quadrupling the radius, or multiplying it by 4, will increase the velocity by a factor of 42, or 16. Meanwhile, doubling the length will decrease the velocity by a factor of 2. The overall change in velocity is thus 16/2, or an increase by a factor of 8.

Question 36

How many GL are equal to 55,000 cL?

Answer 36

For metric conversions, it is often easiest to use scientific notation. 55,000 cL is 55,000 centiliters, which is the same as 55,000 x 10-2 liters. We can rewrite this in proper scientific notation as 5.5 x 102 liters. 1 GL, or 1 gigaliter, is equal to 1 x 109 liters. We can thus do the following calculations: 5.5 x 102 L/(1 x 109 L) = 5.5 x 10-7 L

Question 37

What is the energy of a photon of light with a wavelength of 785 nm? Note that Planck’s constant is 6.62 x 10-34 m2kg/s.

Answer 37

This answer choice is correct. First, we must convert 785 nm into meters, giving 785 x 10-9 m. We can then turn this into proper scientific notation, which would be 7.85 x 10-7m. Next, we can plug all of our values into E = hc/λ, where h is Planck’s constant, c is the speed of light, and λ is the wavelength in meters. E = ((6.62 x 10-34 m2kg/s)(3.0 x 108 m/s))/(7.85 x 10-7m) We need to round these values before we can do our calculations. Let’s round 6.62 to 7, 3 can stay as 3, and 7.85 becomes 8. This yields: E = ((6.5 x10-34 m2kg/s)(3 x 108 m/s))/(8 x 10-7m) E = (19.5 x 10-26 m3kg/s2 )/(8 x 10-7m) We can then round 19.5 to 20 and divide by 8: E = 2.5 x 10-19 J

Question 38

An engineering student fashions a very large object with the intention of studying buoyancy. The object is cube-shaped with a side length of 5 m. The top half of the cube is composed of aluminum (ρ = 2712 kg/m3), while the lower half is made of dogwood (ρ = 750 kg/ m3). The student places the object in a tank of hexanol (ρ = 811 kg/m3). The student decides to try placing the cube on an underwater scale on the floor of the hexanol tank. Approximately what reading will the scale give for the cube’s weight in hexanol?

Answer 38

Using Fb = ρVg, we know that the buoyant force on the cube is 1 × 106 N. However, we also need to know the weight of the cube in air, for which we need to calculate its mass. Since the volume of the cube is 125 m3, 62.5 m3 is composed of aluminum and 62.5 m3 is made of dogwood. For the aluminum, (62.5 m3)(2712 kg/m3) is approximately (6.0 × 10)(2.7 × 103), or 16.2 × 104 kg. For the dogwood, (62.5 m3)(750 kg/m3) is about (6.0 × 101)(7.5 × 102), or 4.5 × 104 kg. Finally, we can add these values to obtain a total mass of around 20.7 × 104, or 2.1 × 105, kg. This yields a weight of 2.1 × 106 N. Last of all, the apparent weight of the cube in hexanol is the difference between the actual weight and the buoyant force, or 1.1 × 106 N.

Question 39

What is the percent composition by mass of copper in a covered beaker containing equimolar amounts of copper (II) nitrate (molar mass 187.6 g/mol), sulfur (molar mass 32.06 g/mol), water, and nitrogen monoxide? Assume none of the species react in the beaker.

Answer 39

Since the question states that equimolar amounts of all four substances are present, we can simply take the mass of copper in one mole of copper nitrate and divide by the sum of the molar masses of these species, including copper nitrate itself. The total of the four molar masses is 187.6 + 32.06 + 18 + 30, or about 267.7. Thus, the percent composition of Cu is equal to (63.5 / 267.7) × 100% = 23.7%. Note that rounding to 60 / 270 makes the math significantly easier and simplifies to 2 / 9, or 22%.

Question 40

Neil wants to increase the intensity of his music by 100-fold. To do this, Neil should increase the intensity by how many decibels?

Answer 40

Increasing the volume by 100 fold is equivalent to increasing the volume by 102. To convert into decibels, multiply exponent by 10. In this example, the exponent is 2 and so 2 x 10 = 20 decibels. dB = 10 log (I/I0) dB = 10 log (100) dB = 10 \* 2 dB = 20