GC Practice Quiz Problems Ch. 7/8/9

Predict if the following are soluble or insoluble in water: -AgCl -Na2CO3 -Sr(OH)2 -AlPO4 -Zn(CH3COO)2

-AgCl: insoluble (s) -Na2CO3: soluble (aq) -Sr(OH)2: insoluble (s) -AlPO4: insoluble (s) -Zn(CH3COO)2: soluble (aq)

Predict if the following are soluble or insoluble in water: -AgNO3 -PbI2 -(NH4)2CO3 -HgCl2 -MgS

-AgNO3: soluble (aq) -PbI2: insoluble (s) -(NH4)2CO3: soluble (aq) -HgCl2: insoluble (s) -MgS: insoluble (s)

Ammonia, NH3(g) and oxygen, O2(g) were allowed to react until equilibrium is established according to the reaction 4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g) ∆H°= -904 kJ Which of the following will cause the equilibrium to shift to the left, decreasing product yield? A. increasing the temperature B. increasing the container volume C. adding a catalyst D. adding more O2(g) E. removing some NO(g)

A. Negative ∆H means exothermic rxn, gives off heat aka heat=product. Therefore, increasing temperature is increasing product, shifting the equilibrium to the left.

Of the following sulfide salts, _______has the lowest molar solubility in water. A. CuS, Ksp=8e-34 B. CdS, Ksp=1e-24 C. NiS, Ksp=3e-16 D. ZnS, Ksp=2e-22 E. PbS, Ksp=3e-25

A. The solubility of salt is determined by its solubility product constant, Ksp. For salts with the same number of cations and anions in its formula, we can use Ksp to determine the relative solubility. The higher the Ksp, the greater the amount of ions in solution which means the higher the solubility of the salt is. For the type of salts given, the solubility can be calculated from the Ksp expression. For example, for CuS, the pertinent equilibrium reaction is CuS(s) ⇌ Cu2+(aq) + S2-(aq) and the Ksp expression is Ksp = [Cu²⁺][S²⁻] = s∙s = s² where s is the solubility. From this equation, we see that s = sqrt(Ksp). Looking at our answer choices, [A] has the smallest Ksp, and therefore the lowest solubility.

The solubility of BaSO4 in water is 2 mg/L at 25 °C. A 2 mg BaSO4 sample was added to 100 mL water. Some of the solid dissolved while 1.8 mg of it remained as precipitate. The solution is said to be_______. A. saturated B. supersaturated C. unsaturated D. dilute E. pure

A. saturated A saturated solution is one that contains the maximum amount of solute that a solvent can dissolve. In such a solution, the rate of dissolution is equal to the rate of crystallization. If the amount of solute dissolved is greater than the solubility, then we have a supersaturated solution. On the other hand, if the solute dissolved is less than the solubility, then an unsaturated solution is formed. Thus, the solution being described is a saturated solution. When 2 mg BaSO4 is added to 100 mL, only 0.2 mg of it dissolved (corresponding to a solubility of 0.2 mg/100 mL = 2 mg/L) and the rest remained as solid.

Arrange the following alcohols in order of increasing solubility in water. I. methanol (CH3OH) II. ethanol (CH3CH2OH) III. n-propanol (CH3CH2CH2OH)

All have 1 O-H bond, so compare by counting number of carbons. More C-H means more nonpolar, and thus, less soluble in water. n-propanol has the most carbons, so it is the least soluble. Methanol has the least carbons, so it's the most polar, and thus most soluble. (III<II<I)

In which of the following solvents is carbon tetrachloride, CCl4, most soluble into, and what intermolecular force between solute and solvent is the most important? A. water, H-bonding B. hexane (C6H14), dispersion C. acetone (CH3C(O)CH3), dispersion D. ethanol (C2H5OH), dipole-dipole E. chloroform (CHCl3), dispersion

B.

Arrange the following in order of increasing ∆G° (most negative to least negative) at 298 K. I. N2(g) + 3 H2(g)→2 NH3(g) K = 1.45e-5 II. H2(g) + I2(g)→2 HI(g) K = 50.5 III. PCl5(g)→PCl3(g) + Cl2(g) K = 0.50 A. I < III < II B. II < III < I C. III < I < II D. I < II < III E. III < II < I

B. II<III<I ∆G°=-RTlnK From the above equation, we see that the larger the value of K, the smaller (more negative) the value of ∆G° is. Also note that for K > 1, lnK>0, so ∆G° is negative, which implies a spontaneous (product-favored) reaction.

Which one of the following is most likely to dissolve in toluene, C7H8? A. HBr B. NaCl C. C6H14 D. CH3CN E. CHCl3

C.

Which of the following will increase the solubility of FeS? I. increasing the temperature II. decreasing the pH III. adding Na2S A. I only B. II and III only C. I and II only D. I and III only E. I, II, and III

C. Increasing the temperature increases the solubility of salts. This is due to an increase in the kinetic energy as the temperature is increased. Decreasing the pH, i.e. making the solution acidic, will increase the solubility of FeS. This is because S2- is weakly basic, and reacts with an acid, thereby decreasing its concentration. Le Chatelier's predicts that the equilibrium shift to the right, increasing solubility. Adding more Na2S will decrease the solubility of FeS, due to the presence of a common ion (S2-), which shifts the equilibrium to the reactant side.

Which of the following statements is true regarding a first order reaction? A. The half-life of a first order reaction increases with increasing initial concentration of the reactant. B. The half-life is dependent of the initial concentration of the reactant. C. The half-life is independent of the initial concentration of the reactant. D. The rate constant of a first order reaction increases with time. E. The rate of a first order reaction increases with time.

C. Remember, half life represents the time it takes for a substance to decrease to one half the initial concentration. For a first order reaction with reactant A, rate = k[A]. The half-life for a first order reaction is given as t1/2=ln2/k. Since there is no concentration term in the equation above, the half-life is independent of the concentration of reactant. The rate constant only varies with temperature. It doesn't matter what the concentration of the reactant is, the half-life time is constant. Since the concentration of [A] decreases with time, so does the rate.

Why is methanol more soluble than isopentyl alcohol (which is also known as isoamyl alcohol)? A. Methanol has less steric hindrance B. Water can more directly solvate methanol C. Methanol has fewer hydrocarbons D. Branched alkanes aggregate to form precipitates E. None of the above. Methanol is less soluble than isopentyl alcohol.

C. Alcohols, having an -OH group, are capable of hydrogen bonding with water, facilitating its solubility in water. Their hydrocarbon portion is nonpolar thereby decreasing their solubility. Thus, the solubility of alcohols decreases as we add more hydrocarbons.

All of the following would triple the osmotic pressure, EXCEPT A. Temperature increase from 298K to 894K B. Changing the solute from 0.2M acetic acid to 0.2M CaCl2 C. Changing the solute from 0.2M NaCl to 0.6M Na3PO4 D. Tripling the amount of NaCl solute in a solution

C. Osmotic pressure is the pressure that must be applied to prevent net movement of water from lower concentration to higher concentration. The osmotic pressure, Π, of a solution is given by Π = iMRT where i is the van't Hoff factor, M is the molarity of the solution, R is the ideal gas constant, and T is the absolute temperature. The van't Hoff factor is determined by the number of particles that the solute dissociates into in solution (i = 1 for nonelectrolyte). Tripling the temperature (A), tripling the van't Hoff factor (B), or tripling the molarity (D) would triple the osmotic pressure. In C), we double the van't Hoff factor and also triple the molarity, which would result in six times the osmotic pressure.

Of the following aqueous solutions, _____has the lowest freezing point. A. 0.20 m acetic acid B. 0.20 m sucrose C. 0.20 m NaCl D. 0.20 m NH4OH E. they all freeze at the same temperature

C. The freezing point of solvent is decreased when added with a nonvolatile solute. This is because the vapor pressure of the solution is always less than that of the solvent. The freezing point depression. DTf, can be expressed mathematically as ΔTf = imKf where i is the van't Hoff factor (related to how many particles the solute dissociates into in solution, 1 for nonelectrolytes) m is the molality, and Kf is the freezing-point depression constant characteristic of the solvent. We are given four solutions with the same concentration. We now compare the van't Hoff factors for each solute. Sucrose is a nonelectrolyte, i.e. it does not dissociate into ions in solution, so its i = 1. Acetic acid, NaCl, and NH4OH all dissociate into ions in solution. Acetic acid and NH4OH are weak electrolytes, they are only partially dissociated into ions in solution. NaCl, on the other hand, is a strong electrolyte, i.e. it completely dissociates into ions in solution. So NaCl has the highest i value and thus the lowest freezing point.

The reaction A → B follows first-order kinetics and has a half-life of 100 s. How long would it take for the concentration of A to decrease from 1.0 M to 0.125 M? A. 100 s B. 200 s C. 300 s D. 400 s E. 500 s

C. 300 s If one half-life is 100 s, then the concentration decrease from 1.0 M to 0.5, then 0.5 to 0.25 then finally 0.25 to 0.125 M would take 3 half-lives, and thus 300 seconds.

Which of the following solutes would decrease water's freezing point by the greatest amount, per mole of solute added? A) NaCl B) NH4NO3 C) NH4Cl D) Na2SO4 E) Sucrose

Compare Van't Hoff factors. NaCl dissolves and dissociates into 2 ions, therefore i=2. NH4NO3: i=2; NH4Cl: i=2; sucrose dissolves but doesn't dissociate, so i=1. Na2SO4 dissociates into 2 Na and 1 SO4, so i=3. D) Na2SO4 would decrease FP by the most.

Consider the reaction at equilibrium 3 Fe(s) + 4 H2O(g) ⇌ Fe3O4(s) + 4 H2(g) Which of the following would happen when more H2O(g) is added to the system at equilibrium? A. The system will adjust by shifting the equilibrium to the reactant side. B. The amount of Fe(s) will increase C. The equilibrium constant, Keq will increase. D. The amount of H2(g) will increase E. Adding H2O will have no net effect.

D. Le Chatelier's principle predicts that when additional H2O (g) is added, the system will adjust to decrease the concentration of H2O (g), so the equilibrium will shift towards the product side, increasing the amounts of Fe3O4 and H2. The equilibrium constant will not be affected by changes in concentration. Only temperature can affect the value of the equilibrium constant.

Determine the vapor pressure of a solution of chloroform and dichloromethane, where the mole fraction of chloroform is 0.40. The vapor pressure of chloroform is 160 mmHg, while that of dichloromethane is 540 mmHg. A. 160 mmHg B. 350 mmHg C. 380 mmHg D. 388 mmHg E. 540 mmHg

D. Ptotal=(0.4*160)+(0.6*540)=64+324 =388 mmHg

Consider the following reaction: 2 A + B → 3 C What is the rate of formation of C at an instant when A is reacting at a rate of 0.60 M/s? A. 0.10 M/s B. 0.20 M/s C. 0.40 M/s D. 0.90 M/s E. 1.80 M/s

D. rate=-∆[A]/2∆t=-∆[B]/∆t=∆[C]/3∆t -∆[A]/∆t=0.60 M/s What is ∆[C]/∆t? From the balanced formula, we can see that for every 2 A that react, 3 C are formed. Thus, ∆[C]/∆t=(3/2)*0.60 M/s=0.9 M/s

Chloroform dichloromethane boils at 60 °C and 40 °C, respectively. Equimolar amounts of chloroform and dicholoromethane were mixed. The resulting solution will boil_____. A. above 61 °C B. below 40 °C C. at 60 °C D. at 50 °C E. at 40 °C

D. Boiling point is defined as the temperature at which the vapor pressure equals the atmospheric pressure. Since the vapor pressure of the solution is somewhere in between the vapor pressure of the pure liquids, then we expect the boiling point of the solution to be in between the boiling points of the pure liquids. If equimolar amounts were not mixed (ie. more of one of the solutions) the boiling point of the solution will be closer to the boiling point of the component with the higher mole fraction.

A concentrated solution of nitric acid contains 90% HNO3 by mass and has a density of 1.50 g/mL. A. 1 mL of this solution contains 1.50 g of HNO3 B. 100 g of this solution contains 150 g of HNO3 C. 1 L of this solution contains 90 g of HNO3 D. 100 g of this solution contains 90 g of HNO3 E. 1.50 mL of this solution contains 90 g of HNO3

D. The percent by mass of a solution is the mass of solute dissolved in 100 g of solution (mass of solute+mass of solvent). %mass = (g solute/g sol'n) × 100% Thus, a 90% HNO3 mass means that there are 90 g of HNO3 dissolved per 100 g of solution. The density of a solution is the mass of solution per mL of solution. A 1.50 g/mL solution means the solution weighs 1.50 g for every mL of solution.

Of which of the following salts is the solubility higher in 0.10 M HCl solution than in pure water? I. AgCl II. BaF2 III. PbI2 A. I only B. II and III only C. I and II only D. II only E. I, II, and III

D. II only Salts whose anions are basic (anions that are the conjugate base of a weak acid are basic) will be more soluble in acidic solution than in a neutral/basic solution. Of the given choices, only BaF2 will be more soluble in 0.10 M HCl, because F- is basic (F- is the conjugate base of the weak acid HF). BaF2(s) ⇌ Ba2+(aq) + 2 F- (aq) In acidic solution, F- will react with the acid (0.10 M HCl solution), thereby decreasing the F-concentration. The system will adjust to increase the concentration of F-, so the equilibrium will shift to the right, increasing the solubility of BaF2. HCl and HI are both very strong acids, and therefore the anion salts are very weakly basic

The reaction 2 A + B → A₂B is zero order with respect to A and second order with respect to B. Doubling the concentration of A and increasing the concentration of B by a factor of three would_____. A. double the rate B. increase the rate by a factor of six C. decrease the rate by half D. decrease the rate by 1/6 E. increase the rate by a factor of nine

E. The rate law for the reaction can be written as Rate = k[A]⁰[B]²= k[B]² Thus, doubling the concentration of A will not affect the rate, while increasing the concentration of B by a factor of three will increase the rate by a factor of nine.

Consider the reaction 2 NO(g) + O2(g) → 2 NO2(g) Suppose equimolar amounts of NO and O2 were placed in a sealed container and allowed to react. Which of the following statements regarding the reaction is correct? I. The rate of the reaction increases as the reaction progresses. II. Decreasing the volume of the container increases the rate. III. The rate constant increases with time. A. I and II only B. I and III only C. II and III only D. I, II, and III E. II only

E. II only The rate of a chemical rxn depends on the [reactants]. As the rxn progresses, reactants are converted into products, so their concentration decreases. Therefore, the rate decreases with time. Decreasing the volume of the container increases the pressure of gaseous reactants (pressure and volume are inversely related), thereby increasing the reaction rate. The rate constant does not change with time. Only by changing the temperature does the value of the rate constant change.

Which of the following statements is true regarding chemical equilibrium? I. The concentrations of reactants and products at equilibrium are constant, which means that all chemical processes have ceased at equilibrium II. The concentrations of reactants and products are always equal at equilibrium. III. The equilibrium constant will change as temperature changes. A. I and III only B. II and III only C. I and II only D. II only E. III only

E. III only

A mixture of Br2 and Cl2 was allowed to react in a closed container and attain equilibrium according to the reaction Br2(g) + Cl2(g) ⇌ 2BrCl(g) Which of the following changes would increase the yield of the reaction? I. increase the container volume II. add more Br2(g) III. remove some Cl2(g)

II only

The reaction N2(g) + 3 H2(g) ⇌ 2NH3(g) has an equilibrium constant Kc = 40 at 298 K. Which reaction mixture proceeds to the right to achieve equilibrium? Reaction I: 1 mol N2, 1 mol H2, 7 mol NH3 Reaction II: 4 mol N2, 1 mol H2, 2 mol NH3 Reaction III: 3 mol N2, 5 mol H2, 0 mol NH3

If Q<K, then reaction goes to the right to achieve equilibrium. By plugging into the K expression (Kc=[NH3]²/[N2][H2]³) the molar values for each component in the reaction mixture for reactions I, II, and III: Q for rxn I=49 Q for rxn II=1 Q for rxn III=0 Therefore, II and III would proceed to the right to achieve equilibrium.

Nitrogen gas' solubility at ambient temperature and 1 atm of pressure is 6.8e-4 mol/L. If the pressure of nitrogen gas in the air is reduced to 0.5 atm, then what is its new concentration?

If the pressure decreased from 1 atm to 0.5 atm, it decreased by a factor of 2. Therefore, the solubility decreases by a factor of 2 (divide by 2). The new concentration is 3.4e-4 mol/L.

You have 2.0 L of 0.5 M NaCl. How many grams of NaCl do you have?

M=mol solute/L solution 2.0 L NaCl*(0.5 mol/L NaCl)(58.5 g/mol NaCl)=58.5 g NaCl

The reaction NO2 (g) + CO (g) → NO (g) + CO2 (g) has a rate law that is second order overall. What are the units of the rate constant?

M⁻¹∙s⁻¹

Rank the following aqueous solutions in order of increasing boiling point. I. 0.20 M CaCl2 II. 0.30 M sucrose III. 0.40 M NaCl

The boiling point of a solvent is increased in the presence of a nonvolatile solute. The increase in boiling point is determined by the product of the concentration, C and the number of particles in solution, i. I. 0.20*3=0.6 II. 0.30*1=0.3 III. 0.40*2=0.8 II<I<III

Consider the following reaction: 2N₂O₅→4NO₂+O₂ If O₂ is being produced at a rate of 2 M/s (mol/L/s), how fast is NO₂ being produced? If O₂ is being produced at a rate of 2 M/s, what is the rate of change of N₂O₅?

The relative rate expression for this reaction is: Rate=-∆[N₂O₅]/2∆t =∆[NO₂]/4∆t=[∆O₂]/∆t We are given the value of [∆O₂]/∆t=2 M/s, which we can set equal to ∆[NO₂]/4∆t and multiply both sides by 4 to get the rate at which NO₂ is produced, which is 8 M/s. We do the same for -∆[N₂O₅]/2∆t, and multiply by -2 to get -4 M/s.

Which of the following aqueous solutions has approximately the same boiling point as that of a 0.15 m NaCl solution? A) 0.1 m MgCl2 B) 0.20 m ethylene glycol C) 0.15 m glucose D) 0.15 m Na2S E) 0.20 m AlCl3

To find the answer, we can multiply the Van't Hoff factor and molality components of the boiling point elevation equation to see which answer choice matches the i*m of NaCl. -NaCl dissociates into 2 moles of ions, so i=2*0.15m=0.3 -ethylene glycol dissolves but doesn't sep into ions, so i=1*0.2=0.2 -glucose does the same, so i=1*0.15=0.15 -Na2S dissolves into 3 ions, so i=3*0.15=0.45 -AlCl3 dissolves into 4 ions, so i=4*0.20=0.8 -MgCl2 dissolves into 3 ions, so 0.1*3=0.30 (A is the correct answer)

What mass (in grams) of MnSO4 (150 g/mol) is required to prepare a 5.0 m manganese sulfate solution with 500 g of water?

m=mol solute/kg solvent 500 g H2O*(1 kg/1000 g H2O)(5.0 mol Mn SO4/kg H2O)(150 g/mol MnSO4)=375 g MnSO4

Which of the following solutes would change water's boiling point by the greatest amount, per mole of solute added? A) 1.2 m glucose B) 0.8 m Ca(NO3)2 C) 1.0 m KCl

glucose: 1.2*1=1.2 calcium nitrate: 0.8*3=2.4 potassium chloride: 1.0*2=2 B is the correct answer.

If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water's boiling point, if KB=0.512 °C/m? In other words, what is the boiling point elevation?

∆TB=(2)(0.512 °C/m)(1.0 m)=1.024 °C