# Chemistry 102 : Can You Pass This Trivia Test? Quiz

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Catherinehalcomb
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Quizzes Created: 1509 | Total Attempts: 5,237,716
Questions: 30 | Attempts: 228  Settings  Are you taking chemistry 102? Can you pass this trivia test? One of the keys to understanding everything you are taught in chemistry class is to have adequate knowledge on the different chemical elements in existence and the different changes exhibited when they interact and the formulas as well. Take the test and stand a chance at refreshing your memory on what you have learnt so far. All the best!

• 1.

### Calculate the concentration of CO2 , P= 3*10-4 atm , if the K=3.4*10-2 M/atm.

• A.

16*10-7

• B.

10*10-6

• C.

11*10-6

• D.

17*10-7

B. 10*10-6
Explanation
The concentration of CO2 can be calculated using the formula: concentration = pressure / K. In this case, the pressure is given as 3*10-4 atm and the value of K is given as 3.4*10-2 M/atm. Plugging these values into the formula, we get concentration = (3*10-4) / (3.4*10-2) = 8.82*10-3 M. However, the answer provided is 10*10-6, which is not consistent with the calculation. Therefore, the given answer is incorrect.

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• 2.

### If the molar mass of glucose is 180g/mol, the solubility of glucose in the water at 200c is 30g/100ml is a 1.22M solution of glucose at 200c:

• A.

Saturated

• B.

Unsaturated

• C.

Super saturated

• D.

None

B. Unsaturated
Explanation
The given information states that the solubility of glucose in water at 20°C is 30g/100ml. This means that 100ml of water can dissolve 30g of glucose. However, the question asks about a 1.22M solution of glucose at 20°C. Molarity is calculated by dividing the number of moles of solute by the volume of the solution in liters. Since the molar mass of glucose is 180g/mol, a 1.22M solution would require 1.22 moles of glucose in 1 liter of solution. Therefore, the given solubility of 30g/100ml is less than the required amount for a 1.22M solution, indicating that the solution is unsaturated.

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• 3.

### 2M solution of Nicl2 has a density of 2.18Kg/ml, calculate the molality if ( M.m=129.7g/mol)?

• A.

9*10-3 m

• B.

9.175*10-4 m

• C.

8.1*10-4 m

• D.

9.9*10-4 m

B. 9.175*10-4 m
Explanation
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the given solution has a density of 2.18 kg/ml, which means that 2 ml of the solution has a mass of 2.18 kg. Since the molecular weight of Nicl2 is 129.7 g/mol, we can calculate the number of moles of Nicl2 in 2 ml of the solution. Dividing this by the mass of the solvent (2.18 kg), we get the molality of the solution as 9.175*10^-4 m.

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• 4.

• A.

0.71

• B.

0.55

• C.

0.29

• D.

0.77

C. 0.29
• 5.

### Calculate the freezing point and boiling point of solution made of 25% by mass of CH2OHCH2OH (62 g/mol) in water give that (Kf of H2O=1.86) (Kb of H2O=0.5)

• A.

-9.92 , -102.72

• B.

-9.92,102.72

• C.

9.92, 102.72

• D.

9.92, -102.72

B. -9.92,102.72
Explanation
The freezing point and boiling point of a solution can be calculated using the formula:

ΔTf = Kf * m
ΔTb = Kb * m

where ΔTf is the change in freezing point, ΔTb is the change in boiling point, Kf is the freezing point depression constant, Kb is the boiling point elevation constant, and m is the molality of the solution.

In this case, the solution is made of 25% by mass of CH2OHCH2OH (62 g/mol) in water. To calculate the molality, we need to find the moles of CH2OHCH2OH and the moles of water.

Moles of CH2OHCH2OH = (25/100) * (mass of solution / molar mass of CH2OHCH2OH)
Moles of water = (75/100) * (mass of solution / molar mass of water)

Once we have the moles of CH2OHCH2OH and water, we can calculate the molality:

molality = moles of solute / mass of solvent (in kg)

Using the molality, we can then calculate the change in freezing point and boiling point using the given constants Kf and Kb.

The correct answer is -9.92, 102.72.

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• 6.

### 2NO2-----> 2NO+O2 has a rate constant K=0.63 M-1 S-1 , IF [NO2]=0.1 M, what is the half life ?

• A.

15.8

• B.

15.0

• C.

16

• D.

16.1

A. 15.8
Explanation
The given reaction is a first-order reaction, as it follows the rate law rate = k[NO2]. The half-life of a first-order reaction can be calculated using the equation t1/2 = ln(2)/k, where k is the rate constant. In this case, the rate constant is given as 0.63 M-1 S-1. Plugging this value into the equation, we get t1/2 = ln(2)/0.63 = 15.8 seconds. Therefore, the correct answer is 15.8.

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• 7.

### Ea= 65.7 KJ/mol, how many times faster the rxn will occur at 500c to 00c?

• A.

77.32

• B.

88

• C.

88.234

• D.

90

C. 88.234
Explanation
At higher temperatures, the rate of a chemical reaction generally increases. This is because higher temperatures provide more energy to the reactant molecules, allowing them to overcome the activation energy barrier more easily and react faster. The given answer, 88.234, suggests that the reaction will occur approximately 88.234 times faster at 500°C compared to 0°C.

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• 8.

### 2N2O5 ------> 4NO2+O2, and (K=6.82 S-1) we stated with 0.025 mol in 2 L, How many moles of N2O5 will remain after 5 min?

• A.

3.232*10-3 mol

• B.

4*10-2 mol

• C.

5*10-3 mol

• D.

3.99 *10-3 mol

A. 3.232*10-3 mol
Explanation
The given reaction is a decomposition reaction where 2 moles of N2O5 produce 4 moles of NO2 and 1 mole of O2. The rate constant (K) is given as 6.82 S-1.

To find the moles of N2O5 remaining after 5 minutes, we can use the first-order rate equation:

ln([N2O5]t/[N2O5]0) = -Kt

Where [N2O5]t is the concentration of N2O5 at time t, [N2O5]0 is the initial concentration of N2O5, K is the rate constant, and t is the time.

Plugging in the given values, we have:

ln([N2O5]t/0.025) = -6.82 * 5

Solving for [N2O5]t, we get:

[N2O5]t = 0.025 * e^(-6.82 * 5)

Calculating this expression gives us approximately 3.232 * 10^-3 mol of N2O5 remaining after 5 minutes.

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• 9.

### A+B -------> C. if ( V=k[A]n * [B]m ) and [A]=0.1 ,[B]=0.5 ,,, find V and K ? [A] [B] V 0.25 0.25 0.2130 0.25 0.125 0.1065 0.2 0.1 0.0682 0.35 0.1 0.1193 0.175 0.1 0.0596

• A.

V=0.1705 M/S K=3.41 M-1.S-1

• B.

V= 0.77 M/S K = 3.3 M

• C.

V= 1.3 M/S K = 4.1 M-2.S-2

• D.

V=1.1 M/S K=0.5 S-1

A. V=0.1705 M/S K=3.41 M-1.S-1
• 10.

### A catalyst:

• A.

Increase reaction rate by lowering the activation energy of forward reaction

• B.

Increase reaction rate by increasing activation energy for reverse reaction

• C.

Increase the reaction rate by providing an alternative energy

• D.

None of the above

C. Increase the reaction rate by providing an alternative energy
Explanation
A catalyst increases the reaction rate by providing an alternative energy pathway for the reaction to proceed. It lowers the activation energy required for the reaction to occur, allowing more reactant molecules to overcome the energy barrier and form products. This alternative pathway reduces the time required for the reaction to reach equilibrium, thereby increasing the reaction rate.

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• 11.

### (2NO+Br-2 ---------->   2NOBr) KC1=2 ( 2NO- -------> N2+O2) KC2=2.1*1030 At temperature=298 K Calculate Kp of (N2+O2+Br-2à2NOBr)

• A.

4.2*1030

• B.

9.5*10-31

• C.

3.9*10-32

• D.

4*10-32

C. 3.9*10-32
Explanation
The equilibrium constant (Kp) is calculated by multiplying the equilibrium constants of the individual reactions involved in the overall reaction. In this case, the overall reaction is the formation of 2NOBr from 2NO and Br-2. The equilibrium constant for this reaction is given as KC1=2. The other reaction given is the decomposition of 2NO to N2 and O2, with an equilibrium constant of KC2=2.1*1030. To calculate Kp, we multiply KC1 and KC2 together. However, since KC2 is a very large number, it is likely that the product of KC1 and KC2 will be a very small number. Therefore, the correct answer is 3.9*10-32.

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• 12.

### Calculate pH for .5 M of HF (Ka =7.1*10-4)

• A.

1.725

• B.

0.01884

• C.

3.77

• D.

4.4

A. 1.725
Explanation
The correct answer is 1.725. The pH of a solution can be calculated using the formula pH = -log[H+]. In this case, HF is a weak acid and will partially dissociate in water to form H+ ions. Since the concentration of HF is given as 0.5 M, we can assume that the concentration of H+ ions is also 0.5 M. Taking the negative logarithm of 0.5 gives a pH value of 1.725.

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• 13.

### A substance that is capable of acting as both acid and base is called:

• A.

Conjugated

• B.

Amphoteric

• C.

Autocratic

• D.

None

B. Amphoteric
Explanation
An amphoteric substance is capable of acting as both an acid and a base. It can donate a proton (act as an acid) or accept a proton (act as a base) depending on the reaction conditions. This property allows it to react with both acidic and basic substances, making it versatile in various chemical reactions.

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• 14.

• A.

0.1425

• B.

0.77

• C.

0.16875

• D.

13.23

D. 13.23
• 15.

### Which of the following correctly relates the two equilibrium constant for two reaction shown. .. (NOCl-------->NO +1/2 Cl2) K1. (2No+Cl2 -------> 2Nocl ) K2

• A.

K2=K1

• B.

K2=1/(K1)2

• C.

K2=K1

• D.

None

B. K2=1/(K1)2
Explanation
The given answer states that K2 is equal to 1/(K1)2. This means that the equilibrium constant for the second reaction (K2) is equal to the inverse square of the equilibrium constant for the first reaction (K1). This relationship suggests that the second reaction is the reverse of the first reaction, with the stoichiometric coefficients squared. This is a valid explanation for the given answer.

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• 16.

### Which of the following compounds doesn’t show buffer:

• A.

NH3 / NH+4

• B.

HCO-3 / H2CO3

• C.

NaOH / OH-

• D.

None

C. NaOH / OH-
Explanation
NaOH is a strong base and OH- is its conjugate base. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. In order for a compound to act as a buffer, it needs to have both a weak acid and its conjugate base (or a weak base and its conjugate acid). NaOH and OH- do not have a weak acid-conjugate base pair, so they do not show buffer properties.

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• 17.

### At freezing point of water 0c0 when Kw = 1.2*10-15 , calculate [H+] and [OH-] for natural solution:

• A.

1.2*10-15

• B.

3.46*10-8

• C.

2.6*10-7

• D.

None

B. 3.46*10-8
Explanation
At the freezing point of water, the concentration of H+ and OH- ions are equal. Since Kw = [H+][OH-], we can use this equation to calculate the concentration of H+ or OH-. Given that Kw = 1.2*10^-15, we can rearrange the equation to solve for [H+] or [OH-]. In this case, the concentration of H+ is 3.46*10^-8.

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• 18.

### Calculate PH if [OH] = 1.9*10 -6

• A.

5.72

• B.

8.28

• C.

5.26

• D.

None

B. 8.28
Explanation
The pH is a measure of the acidity or basicity of a solution. It is calculated using the formula pH = -log[H+]. In this case, we are given the concentration of hydroxide ions [OH-]. Since water is a neutral solution, the concentration of hydroxide ions is equal to the concentration of hydrogen ions [H+]. Therefore, we can use the given concentration of [OH-] to calculate the concentration of [H+]. By taking the negative logarithm of the concentration of [H+], we can determine the pH of the solution. In this case, the pH is calculated to be 8.28.

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• 19.

### Calculate Ka of NH4+ (Kb of NH3= 1.8*10-5)

• A.

10-4

• B.

5.6*10-10

• C.

1.8*10-7

• D.

2*10-3

B. 5.6*10-10
Explanation
The Ka of NH4+ can be calculated using the Kb of NH3 and the equation Kw = Ka * Kb. Since Kw is a constant value (1.0 * 10^-14 at 25°C), we can rearrange the equation to solve for Ka. By substituting the given value of Kb (1.8 * 10^-5) into the equation and solving for Ka, we get the answer of 5.6 * 10^-10.

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• 20.

### The mixture formed by adding 10 ml of .1 M HBr with 20 ml of .2 M HCl , calculate PH :

• A.

0.1666

• B.

0.778

• C.

0.13

• D.

0.114

B. 0.778
Explanation
The pH of a solution is a measure of its acidity or alkalinity. In this case, we are adding two acidic solutions together, HBr and HCl. The concentration of HBr is 0.1 M and the concentration of HCl is 0.2 M. Since HCl has a higher concentration, it will contribute more to the overall acidity of the solution. The pH of the resulting mixture will be closer to the pH of HCl, which is around 0.778.

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• 21.

### Classify each salt into (acidic, basic, natural) respectively: ( Nacl , CH3NH3Br , CH3COONa )

• A.

Natural, acidic, basic

• B.

Acidic, basic, natural

• C.

Natural, basic, acidic

• D.

None

A. Natural, acidic, basic
Explanation
The correct answer is "Natural, acidic, basic." NaCl (sodium chloride) is a natural salt that does not have any acidic or basic properties. CH3NH3Br (methylammonium bromide) is an acidic salt because it can release H+ ions in solution. CH3COONa (sodium acetate) is a basic salt because it can release OH- ions in solution.

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• 22.

### How many moles of NH4Cl must be added to 2 L of .1 M NH3 to form a buffer solution of PH=9 (Kb NH3= 1.8*10-5)?

• A.

9.25

• B.

0.18

• C.

0.36

• D.

0.22

C. 0.36
Explanation
To form a buffer solution with a pH of 9, we need to have equal concentrations of the weak base, NH3, and its conjugate acid, NH4+. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). In this case, NH3 acts as the base (A-) and NH4+ acts as the conjugate acid (HA). The pKa of NH4+ can be calculated using the equation pKa = -log(Ka), where Ka is the acid dissociation constant. Given that Kb NH3 = 1.8*10^-5, we can use the relationship Kw = Ka * Kb to find the value of Ka. Since Kw = 1.0*10^-14, we can solve for Ka and then find pKa. Using the Henderson-Hasselbalch equation with a pH of 9 and pKa, we can solve for the ratio [A-]/[HA]. Finally, we can use the concentration of NH3 (0.1 M) to calculate the concentration of NH4+ needed to achieve the desired ratio, which is 0.36 moles.

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• 23.

### Calculate PH of .3 M NH3 and .36 NH4Cl , what is PH after adding 80 ml of buffer solution PKa = 9.25 (respectively)

• A.

9.17 / 9.20

• B.

9.20 / 9.17

• C.

2.70 / 2.9

• D.

None

A. 9.17 / 9.20
Explanation
The correct answer is 9.17 / 9.20. This is because the given solution is a buffer solution formed by mixing NH3 and NH4Cl. NH3 is a weak base and NH4Cl is its conjugate acid. The pKa value of 9.25 indicates that the solution is slightly basic. When 80 ml of the buffer solution is added, the pH will remain close to the initial pH of the buffer solution, which is 9.17 / 9.20.

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• 24.

### We have the following equilibrium (2SO3 -------> 2SO2 + O2 ) what will happen to [SO3] if the temperature is increased:

• A.

Increase

• B.

Decrease

• C.

Not change

• D.

Cannot tell

D. Cannot tell
• 25.

### PCl5 ---> PCl3 + Cl2 if Kc =.497 if the initial concentration of PCl5 = 1.66, calculate equilibrium concentration for (PCl5 / PCl3 / Cl2) (respectively):

• A.

0.96 / 0.69 / 0.69

• B.

0.69 / 0.69 /0 .69

• C.

0.69 / 0.97 /0 .97

• D.

None

A. 0.96 / 0.69 / 0.69
• 26.

### Calculate PH if 1.52g of HNO3 (mm=63) in 575 ml:

• A.

0.042

• B.

1.38

• C.

1.333

• D.

1.45

B. 1.38
Explanation
To calculate the pH, we need to determine the concentration of HNO3 in moles per liter (M). First, we convert the mass of HNO3 to moles by dividing it by the molar mass. 1.52g of HNO3 is approximately 0.024 moles. Then, we calculate the molarity by dividing the moles by the volume in liters. The volume is given as 575 ml, which is 0.575 liters. The molarity is therefore 0.024 moles / 0.575 liters = 0.0417 M. Finally, we calculate the pH using the formula pH = -log[H+]. Since HNO3 is a strong acid and dissociates completely in water, the concentration of H+ ions is equal to the molarity of HNO3. Taking the negative logarithm of 0.0417 gives a pH value of 1.38.

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• 27.

### The Kc of the following reaction is ( pbcl2 (s) -----> pb+2 (aq)+ 2cl-(aq) )

• A.

Kc = [cl-] [pb+2] /[pbcl2]

• B.

Kc = [cl-] [pb+2]

• C.

None

B. Kc = [cl-] [pb+2]
Explanation
The correct answer is Kc = [cl-] [pb+2]. This is because the expression for the equilibrium constant (Kc) for a reaction is determined by the stoichiometry of the reaction. In this case, the stoichiometry of the reaction shows that the concentration of the chloride ions ([cl-]) and the lead ions ([pb+2]) are both included in the equilibrium constant expression, while the concentration of the solid lead chloride (pbcl2) does not appear in the expression. Therefore, the correct expression for Kc is Kc = [cl-] [pb+2].

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• 28.

### If PH =2.9 and Ka= 1.8*10-5 calculate [HA]

• A.

1.26

• B.

0.88

• C.

1.88

• D.

1.78

B. 0.88
Explanation
The concentration of [HA] can be calculated using the formula [HA] = (Ka * [H+]) / (10^-pH). Given that Ka = 1.8 * 10^-5 and pH = 2.9, we can substitute these values into the formula to find [HA]. After calculations, we find that [HA] is approximately 0.88.

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• 29.

### Which of the following statment is true?

• A.

-Heat is given off to the surrounding in  an Endothermic process

• B.

-/\H is always nigative in Exothermic process

• C.

-work must be positive  in an  exothermic pricess

• D.

- Heat is gained by the system in an endothermic process .

D. - Heat is gained by the system in an endothermic process .
Explanation
In an endothermic process, heat is gained by the system from the surrounding. This means that the system absorbs heat from the surrounding, resulting in an increase in the system's temperature. This is the opposite of an exothermic process, where heat is released from the system to the surrounding. Therefore, the statement "Heat is gained by the system in an endothermic process" is true.

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• 30.

### A particular chemical reaction involves a single reactant. What is the order of the reaction if the rate increases by a factor of eight when the concentration of the reactant is doubled?

• A.

1st order

• B.

2nd order

• C.

3rd order

• D.

Zero order Back to top