Phases and gases

During the melting process, as heat is added to a solid, the temperature: Question 1 Answer Choices A. slowly decreases. B. never increases. C. sometimes increases. D. always increase.

B. never increases.

An ideal gas fills a closed rigid container. As the number of moles of gas in the chamber is increased at a constant temperature: Question 18 Answer Choices A. pressure will increase. B. the effect on pressure cannot be determined. C. pressure will decrease. D. pressure will remain constant.

A. As the number of moles of gas in the chamber is increased at a constant temperature pressure will increase. If V, R, and T are constants, then the ideal gas law, PV = nRT, implies that P is proportional to n. Therefore, increasing n will cause an increase in P.

Heating a 3.5 gram sample of unknown liquid at atmospheric pressure generates the heating curve below. What is the most likely identity of the compound? Question 11 Answer Choices A. Ammonia (Cp, gas = 2.2 kJ/kg·K; ∆Hvap = 1370 kJ/kg) Correct Answer (Blank) B. Propane (Cp, gas = 1.7 kJ/kg·K; ∆Hvap = 356 kJ/kg) C. Ethanol (Cp, gas = 2.3 kJ/kg·K; ∆Hvap = 840 kJ/kg) D. Water (Cp, gas = 1.9 kJ/kg·K; ∆Hvap = 2260 kJ/kg)

A. Heating a 3.5 gram sample of liquid ammonia at atmospheric pressure generates the heating curve below. The heating curve shows the heating of a liquid from -53°C to -33°C, where it vaporizes, and then shows an increase in the temperature of the gas phase. We are provided only with the enthalpies of vaporization and the specific heats for each compound in the gaseous state. As we have no information about the temperature change or energy input in the gaseous phase, nor are the specific heats for the liquid phases provided, we must determine the identity using the latent heat of vaporization. Given we have 3.5 grams of sample and vaporization required ~4,800 J, the heat of vaporization for our unknown is 4,800 J/3.5 g = 1,370 J/g (ammonia is the unknown and our solution). Note that water can be eliminated based upon the boiling point of the unknown.

Pressure is increased on a sample of supercritical fluid. Given a high enough pressure increase, this can cause: Question 3 Answer Choices A. no change in phase B. a phase change into the gas phase. C. a phase change into the solid phase. D. a phase change into the liquid phase.

A. Pressure is increased on a sample of supercritical fluid. Given a high enough pressure increase, this can cause no change in phase. The critical point is the temperature and pressure beyond which it becomes impossible to distinguish between the gaseous and liquid pases. At pressures and temperatures above this point, the substance is termed a supercritical fluid

A flask contains a mixture of three gases: oxygen, nitrogen, and carbon dioxide. The pressure exerted by N2 is 330 torr, and that exerted by CO2 is 250 torr. If the total pressure of the gases is 725 torr, what is the percent pressure of O2? Question 16 Answer Choices A. 20% B. 32% C. 40% D. 15%

A. The percent pressure of O2 is 20%. The pressure due to the nitrogen and carbon dioxide gases is 330 + 250 = 580 torr. Since the total pressure is 725 torr, the partial pressure of oxygen is the difference: 725 — 580 = 145 torr. Since 145 is 1/5 of 725, the percent pressure of O2 is 1/5 = 20%.

When is the reaction quotient equal to the equilibrium constant? Question 26 Answer Choices A. When ΔG = 0 B. When the forward reaction has stopped C. When the reverse reaction has stopped D. When product concentrations are equal to reactant concentrations

A. A reaction has reached equilibrium once the reaction quotient equals the equilibrium constant (choice A is correct). Forward and reverse reactions do not cease at equilibrium but simply occur at the same rate, resulting in no net change in reactant or product concentrations (choices B and C are incorrect). At equilibrium, the concentration of products and reactants are not necessarily equal (choice D is incorrect).

Increasing reactant concentration increases reaction rate by doing which of the following? Question 4 Answer Choices A. Increasing the probability of a collision B. Increasing the average kinetic energy of the reactants C. Increasing the probability of reaction upon collision D. Stabilizing the transition state

A. Addition of more reactant will increase collision probability, which increases the probability of the reaction occurring (choice A is correct). An increase in the average kinetic energy of molecules in the system would be indicative of an increase in temperature (choice B is wrong) and increasing reactant concentration does not increase the chance of the reaction once a collision occurs, though increasing the temperature would do this too (choice C is wrong). Catalysts stabilize the transition state (something additional reactants are unlikely to accomplish - choice D is wrong).

Which of the following is NOT an assumption of the Kinetic Molecular Theory? Question 6 Answer Choices A. Gas particles have a small, yet finite volume. B. The average kinetic energy of the gas is directly related to the temperature. C. Collisions of gas particles with each other and the walls of the container are elastic. D. The attraction of gas particles to themselves is assumed to be negligible.

A. Gas particles have a small, yet finite volume.

Step 1: NO + Br2NOBr2 (Fast) Step 2: NOBr2 + NO → 2 NOBr (Slow) A chemist determines the rate law of the overall process above to be rate = k [NO][Br2], which seems inconsistent with the given mechanism. What best explains her findings? Question 24 Answer Choices A. An inhibitor was present which affected the first step in the mechanism B. Catalysis of the first step in the mechanism occurred. C. An increased quantity of Br2 was added to the system. D. An increased quantity of NO was added to the system.

A. The chemist finds the first step is now dictating the overall reaction rate. One possible explanation would be the inhibition of the first step, causing it to become the rate determining step (choice A is correct). Catalysis of the first step and increasing reactant concentrations would only further increase the rate of the first reaction, leaving the second step as the rate determining step (choices B-D do not explain the observation and are wrong)

A chemist seals a rigid reaction vessel containing a gaseous reaction at equilibrium and begins pumping in helium gas. She fails to see a change in the relative quantity of reactants and products in the system of study. Which of the following best explains these observations? Question 5 Answer Choices A. Total pressure has increased with no change in the reaction quotient. B. Total pressure has increased with compensatory changes in Q and Keq. C. Total pressure has decreased with no change in the reaction quotient. D. Total pressure has decreased with compensatory changes in Q and Keq.

A. Sealing a rigid container and adding a gas to the vessel will increase total pressure (choices C and D are incorrect). Addition of an inert gas also results in a corresponding decrease in the mole fraction for each gas in the mixture, resulting in no change in partial pressures and therefore no change in Q (choice A is correct and choice B is incorrect). Note that changing reactant or product concentrations cannot change Keq, only a change in temperature changes the value of K.

Atmospheric pressure falls with increasing elevation. Given that the atmosphere comprises 21% oxygen and 78% nitrogen, if the atmospheric pressure is 251 mm Hg at the peak of Mt. Everest, what is the partial pressure of oxygen? Question 6 Answer Choices A. 52 mm Hg B. 160 mm Hg C. 199 mm Hg D. 251 mm Hg

A. The partial pressure of a gas is stipulated by Dalton's Law which states that total pressure of a mixture of gases is equal to the sum of the partial pressures of each constituent. Each partial pressure can then be calculated by multiplying the total pressure by the mole fraction of the gas in question. Thus, at the peak of Everest, (251 mm Hg)(0.21) = 52 mm Hg O2, making choice A correct.

When French fries are cooling on an aluminum sheet, the sheet cools off faster than the fries. This is because: Question 2 Answer Choices A. the specific heat of aluminum is less than the specific heat of water in the fries. B. the heat of vaporization for aluminum is greater than that of the water in the fries. C. the specific heat of aluminum is greater than the specific heat of water in the fries. D. the heat of vaporization for aluminum is less than that of the water in the fries.

A. the specific heat of aluminum is less than the specific heat of water in the fries.

A confined gas at 0°C occupies a volume of 55.6 L. Determine the volume of this quantity of gas if the pressure is held constant but the temperature is lowered to -10°C. Question 10 Answer Choices A. 42.0 L B. 53.6 L C. 5.56 L D. 33.6 L

B. A confined gas at 0°C occupies a volume of 55.6 L. The volume of this quantity of gas will be 53.6 L if the pressure is held constant but the temperature is lowered to -10°C. At constant pressure, the volume drops by a factor of 1/273 for every 1 kelvin (or 1°C) drop in temperature. Thus, the volume here drops by (10 / 273) × (55.6 L) = 2 L, so the final volume is 55.6 - 2 = 53.6 L.

Which of the following chemical processes would have the greatest entropy increase? Question 7 Answer Choices A. Precipitation of a salt B. Evaporation of a liquid C. Condensation of a gas D. Compression of a gas

B. Evaporation of a liquid would have the greatest entropy increase. The question asks us to identify the process where the disorder of the molecules is increasing. Since gases are more disorganized than liquids (i.e., the average molecular spacing is larger), and liquids are more disorganized than solids, the only process among the given choices where entropy increases is evaporation.

Which of the following gases will effuse 4.5 times more slowly than He in an effusion experiment? Question 6 Answer Choices A. F2 B. Kr Correct Answer (Blank) C. Ar D. Ne

B. Kr will effuse 4.5 times more slowly than He in an effusion experiment. According to Graham's law, the ratio of the rates of effusion of two gaseous molecules is equal to the square root of the inverse ratio of their molecular weights: Squaring both sides of this equation gives a molecular weight of the unknown as approximately 81, making Kr the best answer.

Which of the following is NOT a criterion for a gas to be considered ideal? Question 9 Answer Choices A. The molecules have negligible volume. B. The molecules have negligible mass. C. The molecules do not experience intermolecular forces. D. The molecules do not liquefy at low temperature.

B. The molecules having negligible mass is NOT a criterion for a gas to be considered ideal. The most important criteria for an ideal gas are: The molecules do not liquefy at low temperature. The molecules do not experience intermolecular forces. The molecules have negligible volume.

A mixture of gases at 760 torr contains 50% carbon dioxide, 30% helium, and 20% hydrogen by pressure. What is the partial pressure due to carbon dioxide? Question 15 Answer Choices A. 532 torr B. 380 torr C. 152 torr D. 228 torr

B. The partial pressure due to carbon dioxide is 380 torr. If half the mixture is composed of CO2, then half the total pressure is due to CO2. Thus, the partial pressure of CO2 is (1/2)(760 torr) = 380 torr.

A chemist is working with a well-characterized reaction in her laboratory. Addition of an unknown compound to the reaction results in an increase in initial reaction rate. However, upon addition of larger quantities of the same compound, no further increase is detected. What is the most likely identity of the unknown compound? Question 30 Answer Choices A. An intermediate B. A catalyst C. An inhibitor D. A product

B. Addition of a catalyst will increase the rate of a reaction by decreasing the activation energy of the step in question but will not continue to do so with increasing concentrations of catalyst (choice B is correct). While addition of an intermediate may increase the rate of a reaction, the initial rate will not plateau at higher concentrations (choice A is wrong). An inhibitor will likely decrease the reaction rate (choice C is wrong) and addition of a product is unlikely to speed the reaction (choice D is wrong).

Removal of a solid bromine sample from a -80°C freezer results in a phase transition as it warms to room temperature. Which of the following terms best explains this process? Question 16 Answer Choices A. Condensation B. Fusion C. Vaporization D. Deposition

B. Bromine is removed from the freezer as a solid and its temperature will increase until it reaches its melting point. Once there, it melts, a process also known as fusion (choice B is correct), and once completely melted it will continue to increase in temperature until room temperature is reached. Note that bromine is a liquid at room temperature, but this was not necessary information to answer this question as the starting material is a solid, and the only phase transition listed here that begins with a solid is fusion.

Which of the following is the most likely identity of a gas that effuses four times slower than hydrogen? Question 34 Answer Choices A. N2 B. O2 C. Cl2 D. A

B. Graham's law can be used when the relative rates of effusion of two gases are known to find the molar mass of an unknown gas: The unknown gas has a molar mass of 32 g/mol, which matches the molar mass of O2, making choice B the best answer.

The ΔG° of a reaction increases following an increase in temperature. What impact would this have on the equilibrium constant? Question 15 Answer Choices A. It increases. B. It decreases. C. It does not change. D. Inadequate information is provided to determine the outcome.

B. If the ΔG° of a reaction increases with increasing temperature, this will cause a decrease in the equilibrium constant (choice B is correct). Qualitatively, as ΔG° increases, the reaction becomes less spontaneous and favors the reactants to a greater degree. With an increase in reactants at equilibrium, the K decreases since K is a ratio of [products]/[reactants]. Quantitatively, ΔG° = -RT ln K, thus as ΔG° increases, K must decrease.

Which of the following substances, when added to a saturated solution containing both aqueous and solid silver chloride, will cause dissolution of the silver chloride solid through complex ion formation with Ag? Question 18 Answer Choices A. Aqueous silver nitrate B. Aqueous potassium cyanide C. Aqueous sodium chloride D. Aqueous sodium acetate

B. In order to solubilize the solid silver chloride, the aqueous ions must be consumed (Le Châtelier's Principle) causing the equilibrium to shift and dissolve more solid silver chloride. Alternatively, the silver chloride must be converted to a soluble substance, such as a complex ion. Adding neutral salts of silver or chloride is the common ion effect and will actually reduce the solubility of silver chloride, causing more precipitation, not dissolution (eliminate choices A and C). Aqueous potassium cyanide will cause the formation of the complex ion dicyanoargenate, Ag(CN)2- (aq), resulting in the consumption (dissolution) of the solid silver chloride. AgCl(s) + 2 CN-(aq) ? Ag(CN)2-(aq) + Cl-(aq). The addition of sodium acetate will have no effect, since both sodium and acetate salts are known to be completely dissociated in solution (eliminate choice D).

What is the sign of ΔG for the reaction: H2O(s) → H2O(l) at the values of P and T denoted by point A in the following phase diagram? Question 12 Answer Choices A. ΔG is positive B. There is not sufficient information to determine the sign of ΔG. C. ΔG = 0 D. ΔG is negative

C. At point A in the following phase diagram the ΔG for the reaction H2O(s) → H2O(l) is equal to 0. As point A is on the line between solid and liquid, these two phases are in equilibrium with one another. It is always true that ΔG for a reaction at equilibrium is equal to 0.

Inside a half-filled water balloon at 25°C and sea level, the vapor pressure of water is 24 mm Hg. What will the vapor pressure of water in the balloon be if a diver takes it to a depth where temperature is 25°C and pressure is 2 atm? Question 6 Answer Choices A. Cannot be determined from the given information B. 12 mm Hg C. 24 mm Hg D. 48 mm Hg

C. Inside a half-filled water balloon at 25°C and sea level, the vapor pressure of water is 24 mm Hg. The vapor pressure of water in the balloon will be 24 mm Hg if a diver takes it to a depth where temperature is 25°C and pressure is 2 atm. The vapor pressure of a substance depends only on the temperature and the intermolecular forces that substance experiences. In particular, it does not depend on external pressure. Therefore, the vapor pressure of water will not change and 24 mm Hg is the correct answer.

The noble gas xenon can exist in the liquid phase at 200 K, a temperature substantially greater than its normal boiling point, when: Question 18 Answer Choices A. the external pressure is less than the vapor pressure of xenon. B. the temperature is increased slowly. C. the external pressure is greater than the vapor pressure of xenon. D. the external pressure is equal to the vapor pressure of water at STP.

C. The noble gas xenon can exist in the liquid phase at 200K, a temperature substantially greater than its normal boiling point, when the external pressure is greater than the vapor pressure of xenon. The compound will exist as a liquid as long as the external pressure is greater than the compound's vapor pressure. (Once the vapor pressure equals the external pressure, the substance boils.)

What is the ratio of the diffusion rates of hydrogen gas to oxygen gas? Question 14 Answer Choices A. 2 : 1 B. 1 : 2 C. 4 : 1 D. 1 : 4

C. The ratio of the diffusion rates of hydrogen gas to oxygen gas is 4 : 1. Diffusion rate is proportional to the speed of the gas. Since lighter gases travel faster than heavier gases, hydrogen must diffuse faster than oxygen. This eliminates 1 : 4 and 1 : 2. Two gases with the same temperature have the same average molecular kinetic energy. Thus, (1/2) mHvH2 = (1/2) mOvO2, which implies (vH/vO)2 = mO/mH. Since mO/mH = 16, we have (vH/vO)2 = 16, which gives vH/vO = 4/1.

A quantity of an ideal gas takes up 600 cm3 at 27°C. Find its volume at -173°C if its pressure remains constant. Question 17 Answer Choices A. 400 cm3 B. 300 cm3 C. 200 cm3 D. 1800 cm3

C. The volume at -173°C if its pressure remains constant is 200 cm3. If P, n, and R are constants, then the ideal gas law, PV = nRT, implies that V is proportional to T. Since the absolute temperature decreased by a factor of 3 (from 300 K to 100 K), the volume will also decrease by a factor of 3. The new volume is therefore (600 cm3)/3 = 200 cm3.

Which of the following best explains why gases behave less ideally at higher pressures? Question 22 Answer Choices A. Molecules possess greater kinetic energy at higher pressures. B. Molecular collisions become increasingly elastic at higher pressures. C. Molecular interaction occurs to a greater degree. D. As a gas is compressed, its temperature increases.

C. At higher pressures, gaseous molecules are packed in closer proximity to one another and intermolecular forces play a larger role, resulting in non-ideal behavior (choice C is correct). The question stem does not reference changes in temperature, and furthermore an increase in kinetic energy would result in more ideal behavior (choice A is incorrect). An increase in the elasticity of collisions would be characteristic of more ideal behavior, not less (choice B is incorrect). While the compression of a gas does result in an increase in temperature under adiabatic conditions, this does not adequately explain why an increase in pressure decreases ideal behavior (choice D is incorrect).

When 1 mole of K3PO4 is dissolved in 1 L of water, how many moles of ions are produced? Question 3 Answer Choices A. 1 B. 2 C. 4 D. 8

C. Because potassium phosphate is an ionic compound, it is a strong electrolyte and will dissociate in solution (eliminate choice A). The covalent bonds between the P and O will not break: only the ionic bonds will, so choice D can be eliminated. The question asks for the total number of ions formed, not just the number of different types of ions. Since three K+ ions and one PO43- ion will form for every unit of K3PO4, choice C is the better answer.

A chemist wants to make 2.0 L of a 0.50 M K2O solution (mw = 94 g/mol). The proper procedure is to weigh out: Question 14 Answer Choices A. 47 g of K2O in a flask and add 2.0 L of water. B. 47 g of K2O in a flask and add water until the volume of the solution becomes 2.0 L. C. 94 g of K2O in a flask and add water until the volume of the solution becomes 2.0 L. D. 94 g of K2O in a flask and add 2.0 kg of water.

C. Molarity is defined as the number of moles of solute per liter of the total solution. That means the volume of water added to the solute is determined by filling the flask to the desired total volume of the solution, as in choices B or C (eliminating choices A and D). Since 2.0 L of a 0.50 M K2O solution is required, a total of 2.0 L x 0.50 moles/L, or 1.0 mole of K2O is needed. Answer choice C is the best option since the molar mass of the solute is 94 g/mol.

Which variable pair listed below is inversely proportional to each other when all other variables are held constant? Question 3 Answer Choices A. Pressure and temperature B. Volume and temperature C. Moles of gas and temperature D. Moles of gas and pressure

C. Moles of gas and temperature

Which of the following is an alternate measurement for the concentration of a gas? Question 4 Answer Choices A. The temperature of the gas B. None of the other answers are correct C. The partial pressure of the gas D. The quantity of the gas

C. The partial pressure of the gas

CO(g) + NO3(g) → NO2(g) + CO2(g) The overall reaction order for the elementary step above is: Question 10 Answer Choices A. zeroth order. B. first order. C. second order. D. third order.

C. The rate law for this elementary step is rate = k [CO][NO3] which, due to each reactant being first order, is a second order process overall (choice C is correct).

A cook picked up a pan on the stove and, realizing it was too hot, quickly threw it into a nearby pot on the counter filled with water (at 99.9°C). After treating his hand for the burn, he found approximately one liter of water had vanished. Assuming no heat exchange with the surroundings, approximately how much energy did the pan lose in the vaporization of the water? [Note: ΔHvap(water) = 40.6 kJ/mol] Question 11 Answer Choices A. 40 kJ B. 400 kJ C. 2200 kJ D. 4000 kJ

C. This question involves the calculation of the energy required for the vaporization of water. One liter of water, weighing one kilogram, contains 55.5 moles (1000 g / 18 g/mole = 55.5 moles). Thus here we would expect 40.6 kJ/mol * 55.5 moles to result in approximately 2200 kJ of energy required to vaporize one liter of water (choice C is correct). Note that if the water had been at a temperature below boiling, more energy would have been required to heat the solution before vaporization could have occurred.

When solutions of gases and solids dissolved in liquids are heated: Question 5 Answer Choices A. gas solubility decreases and solid solubility decreases. B. gas solubility increases and solid solubility decreases. C. gas solubility decreases and solid solubility increases. D. gas solubility increases and solid solubility increases.

C. gas solubility decreases and solid solubility increases.

Which of the following equations could be used as the basis for determining the relative rates of diffusion of two gases in an enclosed flask once the valve is opened? Question 21 Answer Choices A. F = kQq/r2 B. PE = mgh C. PV = nRT D. KE = (1/2)mv2

D. KE = (1/2)mv2 could be used as the basis for determining the relative rates of diffusion of two gases in an enclosed flask once the valve is opened. Both gases are at the same temperature and therefore will possess the same mean kinetic energy: (1/2)m1v12 = (1/2)m2v22 The relative velocities can then be solved for: v2/v1 = (m1/m2)0.5

According to the kinetic theory, which of the following is NOT true of ideal gases? Question 19 Answer Choices A. All collisions among gas molecules are perfectly elastic. B. There are no attractive or repulsive forces between gas molecules. C. For a sample of gas molecules, average kinetic energy is directly proportional to temperature. D. There is no transfer of kinetic energy during collisions between gas molecules.

D. The following statement is NOT true of ideal gases "there is no transfer of kinetic energy during collisions between gas molecules." The other three choices are all criteria for ideal behavior. Gas molecules are allowed to exchange kinetic energy with each other, though the total energy of the system will not change.

What is the volume of two moles of O2 at STP? Question 13 Answer Choices A. 22.4 L B. 89.6 L C. 11.2 L D. 44.8 L

D. The volume of two moles of O2 at STP is 44.8 L. One mole of an ideal gas occupies 22.4 L at STP. Therefore, two moles would occupy 44.8 L.

A chemist attempts to purify a mixture of NaCl and KNO3 by recrystallization. Based on the information in the table above, which salt would she most likely recover as the precipitate in this process? Question 33 Answer Choices A. NaCl, because it has a higher solubility at low temperatures. B. NaCl, because it has a lower solubility at high temperatures. C. KNO3, because it has a higher solubility at high temperatures. D. KNO3, because it has a lower solubility at low temperatures.

D. A solid needs to be soluble at high temperatures and most importantly insoluble at low temperatures in order to recover it as a precipitate via recrystallization. Choice A should be eliminated since a high solubility at low temperatures means the solid will stay in solution. Choices B and C have no bearing on recovering the solute.

2 NO + O2 → 2 NO2 Which of the following characterizes the rate for the reaction above? Question 7 Answer Choices A. rate = -Δ[NO]/Δt B. rate = -2 Δ[O]/Δt C. rate = -Δ[NO2]/Δt D. rate = -(1/2) Δ[NO]/Δt

D. Reaction rate can be characterized by changes in reactants or products over time. In the generic reaction aA → bB, the reaction rate = -(1/a) Δ[A]/Δt = (1/b) Δ[B]/Δt. Thus the rate expression for the given reaction would be = -(1/2) Δ[NO]/Δt = -Δ[O2]/Δt = Δ(1/2) [NO2]/Δt (choice D is correct).

Which of the following is true of two solutions heated to 70°C? Question 2 Answer Choices A. The same amount of energy was required to raise the temperature of each to 70°C. B. The final temperature following mixing of the two solutions would be 70°C. C. All molecules in each solution are traveling at the same speed. D. The solutions possess the same average molecular kinetic energy.

D. Temperature is a measure of the mean molecular motion in a sample, therefore two samples at the same temperature would possess the same average molecular kinetic energy (choice D is correct). As we do not know the starting temperature, nor the heat capacity of either substance, we cannot determine the amount of energy needed to heat them to 70°C (choice A is incorrect). If the two solutions were mixed, a chemical reaction may occur which could change the final temperature (choice B is incorrect). While the average molecular kinetic energy of the solutions may be the same, this does not mean that every molecule travels at the same speed. Instead, they occupy a distribution of speeds (choice C is incorrect).

SO2(g) + 1/2 O2(g)SO3(g) What is the equilibrium expression for the reaction above? Question 23 Answer Choices A. Keq = [SO2] [O2]1/2/[SO3] B. Keq = [SO3]2/[SO2]2 [O2] C. Keq = [SO3]/[SO2] [O2] D. Keq = [SO3]/[SO2] [O2]1/2

D. The equilibrium constant is dictated by the concentration of products over reactants (eliminate choice A), raised to the power of their coefficients. As each substance is in the gaseous phase (rather than a pure solid or liquid), each species in the reaction is part of the equilibrium expression. Thus for aA + bB cC, we get Keq = [C]c/[A]a[B]b or [SO3]/[SO2] [O2]1/2 (choice D is correct). Note that choice B would be more appropriate if the reaction had been balanced with whole number coefficients instead.

In the phase diagram above, Point x represents a: I. Solid II. Liquid III. Gas Question 2 Answer Choices A. II and III only B. I, II, and III C. I and III only D. I and II only

In the phase diagram above, Point x represents a liquid and a gas, only. Point x lies along the liquid-gas equilibrium curve, so both of those phases will be observed at that point.

A supercritical fluid is a substance that is beyond its critical point and has characteristics of both a gas and liquid. To change a supercritical fluid into a gas, which of the following should be done? Question 5 Answer Choices A. Increase temperature B. Decrease temperature C. Decrease pressure D. Increase pressure

A supercritical fluid is a substance that is beyond its critical point and has characteristics of both a gas and liquid. To change a supercritical fluid into a gas, decrease pressure. Recall that the phase diagram for for most substances (excluding water) looks like the figure below. A supercritical fluid has a temperature and pressure above the critical point. From this diagram the only way to change a supercritical fluid into a gas is to decrease pressure. Decreasing temperature would change the supercritical fluid into a liquid, and increasing either pressure or temperature would have no phase effect on the supercritical fluid.

A 44.8 L container is filled with 1 mole of ideal gas X and 1 mole of real gas Y at 0°C. The pressure inside the container is: Question 8 Answer Choices A. between 0.5 atm and 1 atm. B. 0.5 atm. C. less than 0.5 atm. D. 1 atm.

A. A 44.8 L container is filled with 1 mole of ideal gas X and 1 mole of real gas Y at 0°C. The pressure inside the container is between 0.5 atm and 1 atm. The volume of one mole of an ideal gas at STP is defined as 22.4 L. If both gases were ideal in this 44.8 L container, the 2 moles would give a combined pressure of: However, since gas Y is a real gas, its molecules experience intermolecular forces resulting in less force exerted on the walls of the container. The partial pressure exerted by the real gas is less than its ideal partial pressure. Therefore, the total pressure is 0.5 atm from ideal gas X plus something less than 0.5 atm from real gas Y, making the total pressure between 0.5 atm and 1 atm.

A sample of helium gas at 300°C and 600 torr is in a 200 L container. If the temperature of the gas is raised to 900°C at constant volume, how is the pressure of the gas affected? Question 5 Answer Choices A. Pressure will increase by a factor of 2 since the temperature increases by 600°C B. Pressure will be unaffected in this instance due to the large volume of the container. C. Pressure will decrease by a factor of 3 since pressure and temperature are inversely related. D. Pressure will increase by a factor of 3 since pressure and temperature are directly related.

A. A sample of helium gas at 300°C and 600 torr is in a 200 L container. If the temperature of the gas is raised to 900°C at constant volume, the pressure will increase by a factor of 2 since the temperature increases by 600°C. The choices, "pressure will be unaffected in this instance due to the large volume of the container" and "pressure will decrease by a factor of 3 since pressure and temperature are inversely related" can be eliminated since pressure and temperature are directly related at constant volume. The choice stating "pressure increasing by a factor of 3..." is tempting since the temperature has changed from 300°C to 900°C, a threefold increase. However, when predicting changes for gases, we need to use the absolute temperature, or Kelvin scale. After converting these temperatures to 573 K and 1173 K, respectively, we see that in fact this is roughly a doubling of the absolute temperature.

A sample of water is in equilibrium between the solid, liquid, and gaseous phases. Which of the following perturbations will result in generating only liquid water? Question 10 Answer Choices A. Increasing pressure B. Decreasing pressure C. Increasing temperature D. Decreasing temperature

A. A sample of water is in equilibrium between the solid, liquid, and gaseous phases and increasing pressure will result in the formation of only liquid water. The water sample is initially at the triple point where all three phases are in equilibrium. A decrease in pressure or an increase in temperature would result in the formation of a gas (decreasing pressure and increasing temperature are wrong). Decreasing temperature would result in the formation of solid water (decreasing temperature is wrong). Increasing pressure would result in the formation of a liquid and is the correct answer.

Two substances of equal mass are heated at the same rate for the same period of time without undergoing a phase change. If the specific heat of Substance I is twice that of Substance II, then the change in temperature after heating will be: Question 1 Answer Choices A. twice as large in II as in I. B. the same in I as in II. C. twice as large in I as in II. D. four times as large in II as in I.

A. Two substances of equal mass are heated at the same rate for the same period of time without undergoing a phase change. If the specific heat of Substance I is twice that of Substance II, then the change in temperature after heating will be twice as large in II as in I. The specific heat of a substance is the quantity of heat required to raise the temperature of one gram of the substance by one degree. Therefore, if Substance I has a specific heat which is twice that of Substance II, the temperature of Substance I will only be raised by half as much as Substance II after heating for an equivalent time.

A researcher is working with a sample of neon at a temperature of 285 K and a pressure of 58 atm. If the pressure is lowered to 48 atm, the best prediction would be a phase change from: Question 17 Answer Choices A. gas to liquid. B. liquid to gas. C. liquid to solid. D. gas to solid.

B. A researcher is working with a sample of neon at a temperature of 285 K and a pressure of 58 atm. If the pressure is lowered to 48 atm, the best prediction would be a phase change from liquid to gas. Changing gas to solid, liquid to solid, or gas to liquid would each involve increasing the pressure of a noble gas. The only transformation listed that involves a decrease in pressure is the change from liquid to gas.

A substance is in the gas phase at STP. As the pressure of the surroundings is progressively decreased at constant temperature, the gas will eventually undergo: Question 16 Answer Choices A. sublimation. B. no phase change C. evaporation. D. condensation.

B. A substance is in the gas phase at STP. As the pressure of the surroundings is progressively decreased at constant temperature, the gas will eventually undergo no phase change. Since the substance starts in the gas phase, it cannot evaporate or sublime. A gas can be liquefied by bringing the molecules closer together. This can be accomplished by either increasing the external pressure on the gas or by decreasing its temperature. Neither of these is done here, so this gas will remain a gas.

What is the sign of ΔG for the reaction: H2O (s) → H2O (l) at the values of P and T denoted by point A in the following phase diagram? Question 14 Answer Choices A. ΔG is 0. B. ΔG is negative. C. ΔG depends on the sign of the entropy change. D. ΔG is positive.

B. At point A in the phase diagram, the sign of ΔG for the reaction H2O (s) ? H2O (l) is negative. A phase diagram explains which phase is the lowest energy phase of a material at any given point of P and T. As point A is within the liquid region, the change of any other phase of water to liquid is spontaneous, and has a negative value of ΔG.

Carbon monoxide gas reacts with oxygen to yield carbon dioxide, according to the balanced equation below. If 2.5 mol of O2 reacts with an excess of CO, what is the maximum volume of CO2 that can be produced at 760 mm Hg and 25°C? (R = 0.0821 L·atm/K·mol) 2 CO + O2 2 CO2 Question 4 Answer Choices A. 61 L B. 122 L C. 112 L D. 5 L

B. Carbon monoxide gas reacts with oxygen to yield carbon dioxide, according to the balanced equation below. If 2.5 mol of O2 reacts with an excess of CO, the maximum volume of CO2 that can be produced at 760 mm Hg and 25°C is 122 L. (R = 0.0821 L·atm/K·mol) 2 CO + O2 2 CO2 First, address the stoichiometry of the reaction. For every mole of O2 used, 2 moles of CO2 are produced. Therefore, the question asks for the volume of 5.0 mol CO2 at 1 atm (760 torr) and 298 K (25°C). Use the ideal gas law (the given gas constant R was a big hint) and do some MCAT math approximating, paying attention to the necessary units shown above. PV = nRT or V = nRT / P V = (5 mol)(~0.08 L·atm/K·mol)(~300 K) / (1 atm) V ≈ 120 L

In a mixture of two gases, I and II, at STP, the average velocity of a Gas I molecule is twice that of a Gas II molecule. Among the following, Gases I and II are most likely: Question 20 Answer Choices A. He and Kr. B. Ne and Kr. C. He and Ar. D. Ne and Ar.

B. In a mixture of two gases, I and II, at STP, the average velocity of a Gas I molecule is twice that of a Gas II molecule. Among the following, Gases I and II are most likely Ne and Kr. Two gases having the same temperature have the same average molecular kinetic energy. Therefore, (1/2)mIvI2 = (1/2)mIIvII2, which gives mII/mI = (vI/vII)2. Since vI/vII = 2, we have mII/mI = 22 = 4. Among the available choices, the only gases with a mass ratio close to 4 to 1 are krypton (83.8 g/mol) and neon (20.2 g/mol).

Which of the following statements most clearly defines the P and V terms of the van der Waals equation for real gases? Question 3 Answer Choices A. P is the pressure caused by collisions with the walls of the container and between gas molecules, and V is the volume of the container. B. P is the pressure caused by collisions with the walls of the container, and V is the volume of free space in the container. C. P is the pressure caused by collisions with the walls of the container and between gas molecules, and V is the volume of free space in the container. D. P is the pressure caused by collisions with the walls of the container, and V is the volume of the container.

B. The best definition of the P and V terms of the van der Waals equation for real gases is that P is the pressure caused by collisions with the walls of the container, and V is the volume of free space in the container. Looking at the first part of the answer choices, measurements can only be made to assess the force of collisions of gas particles with the container walls and not the collisions between particles, so choices that indicate both collision types are important can be eliminated. As for the volume part of the answer choices, remember that ideal gas particles don't occupy space. Therefore, the volume of free space in the container is the same as the entire volume of the container. However, since real gas particles have significant volume compared to their container size, the volume in the van der Waals equation refers to the free space within the container that is not occupied by all the particles.

Two gases, each 1 L at STP, are allowed to spontaneously react in an insulated cylinder with a floating piston to form two new gaseous products such that total moles are constant before and after the reaction. Immediately after the completion of the reaction the volume of the cylinder was measured to be 1.82 L. What can be said about the reaction? Question 12 Answer Choices A. The reaction was isothermal. B. The reaction was endothermic. C. ΔS for the reaction was negative. D. The reaction was exothermic.

B. Two gases, each 1 L at STP, are allowed to spontaneously react in an insulated cylinder with a floating piston to form two new gaseous products such that total moles are constant before and after the reaction. Immediately after the completion of the reaction the volume of the cylinder was measured to be 1.82 L, therefore the reaction was endothermic. If the reaction created the same number of moles of gas as it consumed, and the reaction remained at STP, the volume of the cylinder should have been 2 L. Since the eventual volume is less with the pressure equilibrated to atmospheric, the temperature of the reaction must have decreased, as would have been the case if the reaction was endothermic and consumed heat from the system in order to do chemical work. Had the reaction been exothermic, and the reaction given heat to the system, the end volume would have been greater than 2 L. If the reaction were isothermal, at STP, there would have been no change in volume. Finally, ΔS for the reaction could not have been negative, since the number of moles of gas did not change over the course of the reaction.

The following table shows the values of the van der Waals constants for a select group of gases: Based on these data, which of the following statements is true? Question 27 Answer Choices A. The dipole-dipole forces between CO molecules is stronger than the London dispersion forces between CO2 molecules. B. The molecular size of oxygen is smaller than the molecular size of nitrogen. C. Oxygen should have a higher boiling point than carbon monoxide. D. Helium has the largest deviation to ideal behavior at high pressures.

B. Although dipole-dipole interactions are normally stronger than London dispersion forces, the strength of the intermolecular forces for real gases is directly proportional to the value of "a" in the van der Waals equation. Since the value of "a" is actually larger for CO2, this statement is false (eliminate choice A). Molecular size is depicted by the constant "b" in the van der Waals equation. The table shows the value of "b" is larger for nitrogen than for oxygen, so choice B is correct. Boiling point is dependent on the intermolecular forces of the compounds in question, so compare the values of the "a" constant in the table. CO has the larger "a" value, so it has stronger intermolecular forces than oxygen and should boil at a higher temperature (eliminate choice C). Finally, the larger the values of "a" and "b", the larger the deviation from ideal behavior predicted by the ideal gas law. Helium has some of the smallest values of the gases listed, not the largest (eliminate choice D).

Which of the following changes to an equilibrated system is most likely to change the proportion of reactants and products at equilibrium? Question 8 Answer Choices A. Increasing activation energy B. Increasing product stability Correct Answer (Blank) C. Increasing reactant concentration D. Increasing reaction rate

B. Increasing product stability will increase the relative proportion of products at equilibrium (choice B is correct). Changes in reaction rate (and other kinetic changes) are not likely to impact equilibrium (or other thermodynamic properties - choices A and D are wrong). Increasing reactant concentration will temporarily decrease the reaction quotient, Q, but once equilibrium has been re-established, the same proportions of reactants and products will exist (choice C is wrong).

Question 29 To increase the solubility of a gas in solution you could: Question 29 Answer Choices A. stir the mixture. B. cool the mixture. C. move the solution to a higher elevation. D. place it in a container under a vacuum.

B. Stirring a solution has no impact on how much solute can dissolve in a solvent, although it can affect the rate at which a solute (usually a solid) dissolves (eliminate choice A). At higher elevation atmospheric pressure is lowered. Therefore, choices C and D both suggest the same change and must both be incorrect. The solubility of a gas increases as pressure increases, or as temperature decreases, making B the best answer.

Which of the following reactions favors reactants to the greatest degree at equilibrium? Question 31 Answer Choices A. H2(g) + I2(g) ? 2 HI(g)Keq = 2.8 x 101 B. HC2H3O2(aq) ? C2H3O2-(aq) + H+(aq)Keq = 1.79 x 10-5 C. N2O4(g) ? 2 NO2(g)Keq = 8.7 x 10-1 D. 2 NH3(g) ? N2(g) + 3 H2(g)Keq = 2.3 x 101

B. The equilibrium constant is a ratio of products to reactants established at equilibrium. As we are looking for the reaction that favors the reactants to the greatest degree, we are looking for the smallest equilibrium constant (choice B is correct).

Which of the following represents a deviation from the assumptions of the kinetic molecular theory of gases when observed in real gases? Question 1 Answer Choices A. Kinetic energy is conserved upon collisions. B. Particles are instantaneously polarized C. Molecules remain in constant random motion. D. The temperature of the gas depends on its average kinetic energy.

B. The kinetic molecular theory of gases describes the behavior of gases but relies on many assumptions, including: the total volume of gas molecules is negligible, all collisions are elastic (choice A is incorrect), particles are in constant random motion (choice C is incorrect), particles span a distribution of speeds, and gas particles do not interact appreciably (no IMFs). Instantaneous polarization of molecules is characteristic of London dispersion forces and would contribute to deviations from predicted behavior (choice B is correct). Temperature is a function of average kinetic energy of particles and is true of any sample (choice D is incorrect).

When aqueous solutions of AgNO3 and Ba(OH)2 are combined, one should expect to observe: no visible change since all ion combinations create soluble compounds. the formation of a precipitate since AgOH is insoluble. the formation of a precipitate since Ba(NO3)2 is insoluble. Question 25 Answer Choices A. I only B. II only C. III only D. II and III only

B. The reaction suggested is the double displacement: AgNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + AgOH(s). Two common ionic solid solubility rules are that nitrate compounds are soluble, while most Pb, Hg, and Ag solids are insoluble. Because the AgOH forms a precipitate, Roman numeral I is false while Roman numeral II is true. Since all nitrates are soluble, Roman numeral III is also false, making B the best answer.

A 40 g sample of argon at 760 mm Hg and 25°C is cooled to 0°C. What is the final volume of this sample? Question 17 Answer Choices A. 11.2 L B. 22.4 L C. 33.6 L D. 40.0 L

B. This question can be solved using the ideal gas law: V = nRT/P = (1 mol)(0.082 L?atm/mol?K)(273 K) / (1 atm) = 22.4 L. Alternatively, and since the value of the gas constant wasn't given, we know the volume of one mole of gas at STP is 22.4 L. Since a 40 g sample of argon is also one mole, choice B is correct.

Which of the following is NOT a potential advantage to inflating car tires with N2 instead of atmospheric air (79% N2, 20% O2, and 1% other gases including H2O, Ar, and CO2)? Question 7 Answer Choices A. Atmospheric air is more likely to cause corrosion of the tire interior compared to N2. B. Atmospheric air contains water vapor, which is more likely than N2 to condense as the pressure inside the tire increases. C. If a very small hole were poked in the tire, a tire filled with pure N2 would lose pressure less quickly than a tire filled with atmospheric air. D. Tires filled with N2 weigh less than those filled with atmospheric air, leading to better fuel efficiency.

C. If a very small hole were poked in the tire, a tire filled with pure N2 would lose pressure less quickly than a tire filled with atmospheric air is NOT a potential advantage to inflating car tires with N2 instead of atmospheric air (79% N2, 20% O2, and 1% other gases including H2O, Ar, and CO2). If a very small hole is poked in a tire, the gas would leak out in a process approximating effusion. The rate of effusion is inversely proportional to molecular weight according to Graham's law: Since air contains both N2 (MW = 28 g/mol) and O2 (MW = 32 g/mol), oxygen will effuse more slowly. The choice, "tires filled with N2 weigh less than those filled with atmospheric air, leading to better fuel efficiency" is an advantage, since atmospheric air is more dense than pure N2 due to the presence of the heavier O2. It is true that water experiences stronger intermolecular forces than N2, therefore atmospheric air is more likely than N2 to condense as the pressure inside the tire increases. Finally, it is also true that corrosion occurs due to oxidation caused by molecular O2 present only in atmospheric air.

Water boils at a lower temperature at high altitudes. The most likely explanation for this is that at high altitudes: Question 15 Answer Choices A. the vapor pressure of water is increased. B. the temperature is higher than at low altitudes. C. the atmospheric pressure is lower than at low altitudes D. more energy is available to break bonds in liquid molecules.

C. Water boils at a lower temperature at high altitudes. The most likely explanation for this is that at high altitudes, the atmospheric pressure is lower than at low altitudes. A phase change does not involve breaking bonds; intermolecular forces between liquid molecules are overcome during the boiling process. Eliminate the statement "more energy is available to break bonds in liquid molecules". Vapor pressure does not change with a change in pressure, so the statement "the vapor pressure of water is increased" is incorrect. Boiling occurs when the vapor pressure of a liquid equals the atmospheric pressure. Water boils at a lower temperature at high altitudes because the atmospheric pressure decreases as altitude increases.

Which of the following best explains the collision of an ideal gas molecule with the container wall? Question 11 Answer Choices A. There is a net loss of kinetic energy following the collision. B. The gas molecule always increases in speed following the collision. C. The container experiences a change in momentum equal to and opposite that of the molecule. D. The gas molecule always decreases in speed following the collision.

C. When an ideal gas molecule collides with the container wall, the container experiences a change in momentum equal to and opposite that of the molecule. Collisions in an ideal gas are perfectly elastic, meaning that both kinetic energy and momentum are conserved. The speed of a gas molecule following collision depends upon both its initial kinetic energy and that of the container molecule. As both populations will span a distribution of speeds, the gas molecule may gain or lose kinetic energy following collision but will not always increase or decrease in speed.

A researcher places a sample of radiolabeled AgI into a saturated solution of non-radiolabeled AgI and several hours later inadvertently spills some of the liquid on a nearby countertop. Is the researcher in danger while cleaning this spill? Question 13 Answer Choices A. No, the radioactive solid is insoluble so the spilled liquid contained only non-radiolabeled ions. B. No, at equilibrium, the reaction would thermodynamically drive any radioactive iodide into the solid. C. Yes, a portion of the radioactive iodide in the solid would have been replaced by the non-radiolabeled isotope. D. Yes, the iodide initially present in the solid would all enter solution and diffuse equally throughout the liquid.

C. Chemical equilibria are dynamic, so while there is no net flux between reactants and products (the solid and aqueous forms of the solute), this does not preclude the interchanging of individual ions of the two iodide isotopes between these forms (choice C is correct, choices A and B are wrong). Diffusion would play a role in dispersing aqueous iodide, but given we are placing more solid in a saturated solution, some of the radioactive iodide will remain in the solid (choice D is wrong).

A chemist attempts to determine the rate law of a poorly characterized reaction. Propose the best rate law consistent with the following data. A. rate = k [H2O2] B. rate = k [I-] C. rate = k [H2O2] [I-] D. rate = k [H2O2]2 [I-]

C. In the isolation method of determining reaction rate, two trials should be identified where the initial concentration of only one reactant changes at a time so the order of that reactant can then be determined based upon the change observed in the initial reaction rate. Trials 1 and 4 can be compared to determine the order of hydrogen peroxide: given rate = [x]y, 2 = 2y and y = 1. In other words, when the concentration of peroxide increases by a factor of two, the reaction rate increases by the same factor. Therefore the reaction is first order with respect to peroxide. (Incidently, trials 2 and 3 can be used to find the order of peroxide as well.) Similarly, trials 1 and 2 can be compared to determine the order of iodide: given rate = [x]y, 2 = 2y and y = 1. Thus the rate law is rate = k [H2O2] [I-] (choice C is correct).

What is the mole fraction of NaOH in a solution made from 80 g NaOH in 180 g H2O? Question 20 Answer Choices A. 0.31 B. 0.20 C. 0.17 D. 0.11

C. Mole fraction is defined as the number of moles of a given component of a mixture divided by the total number of moles of all components in the mixture. Since the molar mass of NaOH is 40 g/mol, 80 g NaOH is two moles of solute. Similarly, since the molar mass of water is 18 g/mol, 180 g of water is 10 mol of solvent. Therefore, the mole fraction of NaOH can be calculated by 2 mol NaOH / (2 mol NaOH + 10 mol H2O), or 0.17.

A sample of nitrogen gas is cooled from 80°C to 20°C in a rigid-walled container. What impact would this have on pressure? Question 12 Answer Choices A. Increase to 1.2 times the original pressure B. Increase by a factor of 4 C. Decrease to 0.8 times the original pressure D. Decrease by a factor of 4

C. Pressure and temperature increase and decrease proportionally, thus a decrease in temperature results in a decrease in pressure (choices A and B are incorrect). As volume, the number of moles of gas, and the ideal gas constant remain unchanged in this process, we can rearrange the ideal gas law to P1/T1 = P2/T2, or P2/P1 = T2/T1. Remembering to convert temperature to Kelvin when using the gas law equations, we can solve for P2/P1 thus: 293 K / 353K ~ 0.8. The pressure therefore also decreases to roughly 0.8 times the original pressure, making choice C correct.

The van der Waals equation for real gases: contains two constants which have unique values for each gas. When comparing these constants for neon and argon, the constant "a" is larger for argon because: Question 9 Answer Choices A. argon has larger particles than neon. B. argon has more protons than neon. C. argon has stronger intermolecular forces than neon. D. argon has a greater density than neon.

C. The constant "a" is a correction for the attractive forces between the gas molecules. The larger the value of "a," the stronger the intermolecular forces of the gas in question. Only choice C talks about intermolecular forces, so is the correct answer. Choice A, referring to particle size, is dependent on the "b" constant from the equation (eliminate choice A). The number of protons does not have a consistent relationship to intermolecular forces (eliminate choice B). Finally, density (which equals mass/volume) has no correlation to the van der Waals equation (eliminate choice D).

Four identical rigid containers containing equimolar quantities of different gases at identical initial temperature and pressure conditions are subjected to an equivalent amount of heat added via heat gun. Which of the following statements is true of the gas in the container displaying the lowest temperature change? Question 13 Answer Choices A. It is the least dense of the four. B. It has the lowest molar mass of the four. C. It behaves the most ideally of the four. D. It has the largest molar heat capacity of the four.

D. Four rigid containers containing equimolar quantities of different gases at identical initial temperature and pressure conditions are subjected to an equivalent amount of heat added via heat gun. The gas that displays the smallest increase in temperature has the largest molar heat capacity of the four. Heat capacity at constant volume (Δq/ΔT) is a measure of how the temperature of a system changes with the addition or subtraction of heat. A large value of heat capacity implies that a large input of q will result in a small change in T. An ideal gas cannot store added energy in potential forms (rotational, vibrational, etc.), leading to small heat capacities as all added energy will be converted to kinetic energy, and hence increased T. Molar mass isn't a good predictor of heat capacity and can't explain the trend. The least dense gas, since equimolar amounts are used in a given volume, will also be the one with the lowest molar mass, eliminating both these choices.

Which of the following is NOT true concerning the following phase transition diagram of water? Question 4 Answer Choices A. Temperatures of the two flat plateaus are 0°C and 100°C, respectively B. Slope of the segments where temperature changes is inversely proportional to the mass of water C. ΔHvap > ΔHfus D. Slope of the segments where temperature changes is directly proportional to the specific heat of water

D. Given the phase transition diagram of water, it is NOT true that the slope of the segments where temperature changes is directly proportional to the specific heat of water. ΔHvap is always greater than ΔHfus regardless of the substance. The diagram shows this because ΔHfus (the first plateau) is shorter than ΔHvap (the second plateau). The first plateau represents the melting point and the second plateau represents the boiling point of water. The positively sloped portions of the diagram follow the formula q = mcΔT. Therefore, the slope (Δy / Δx) is given by ΔT / Δq = 1 / mc. Mass (m) and specific heat (c) are both inversely proportional to the slope. Also note that each sloped segment will depend on the specific heat of water in its three separate phases: ice, liquid, and steam.

Starting from the conditions at point A on the solid, liquid, gas phase diagram below, which of the following changes will generate a fluid with physical properties resembling both a liquid and a gas? Question 9 Answer Choices A. Decreasing pressure B. Increasing the volume of the container C. Decreasing temperature D. Increasing pressure

D. Starting from the conditions at point A on the phase diagram below, increasing pressure will generate a fluid with physical properties resembling both a liquid and a gas. The situation described where a fluid resembles both a liquid and a gas is known as a supercritical fluid (a fluid beyond the critical point). Point A is directly below the critical point, thus an increase in pressure would generate a supercritical fluid (increasing pressure is correct). Decreasing pressure would maintain the compound in the gas phase. Note that increasing the volume of the container will also result in a decrease in pressure, and both answer choices cannot be correct. A large enough decrease in temperature would pass the compound through its triple point before resulting in the formation of a solid only.

Two blocks of metal, Block 1 and Block 2, have the same mass, but the specific heat of Block 1 is greater than that of Block 2. Assume that the temperature of Block 1 (T1) is less than the temperature of Block 2 (T2). If these blocks are placed in contact and allowed to reach thermal equilibrium, which of the following best describes the temperature change of Block 1 (ΔT1) and the temperature change of Block 2 (ΔT2)? Question 8 Answer Choices A. < B. ΔT2 C. ΔT1 D. ΔT2 < 0 < ΔT1 and

D. Two blocks of metal, Block 1 and Block 2, have the same mass, but the specific heat of Block 1 is greater than that of Block 2. Assume that the temperature of Block 1 (T1) is less than the temperature of Block 2 (T2). If these blocks are placed in contact and allowed to reach thermal equilibrium, the temperature change of Block 1 (ΔT1) and the temperature change of Block 2 (ΔT2) is ΔT2 < 0 < ΔT1 and |ΔT1| < |ΔT2|. Because T1 < T2, heat will flow from Block 2 to Block 1, implying that the temperature change of Block 1 will be positive but the temperature change of Block 2 will be negative. Now, since Block 1 has the greater specific heat, its magnitude of temperature change will be smaller than that of Block 2. These facts are summarized by the inequalities ΔT2 < 0 < ΔT1 and |ΔT1| < |ΔT2|.

When the pressure of a container of a real gas is less than that predicted by the ideal gas law, this deviation is explained by which of the following statements? Question 2 Answer Choices A. The ideal gas law is only valid at very low temperatures and very high pressures. B. The ideal gas law does not correct for the volume of the particles of gas. C. The ideal gas law corrects for the shape of the particle of gas, whereas the real gas law does not. D. The ideal gas law does not correct for the intermolecular forces between the gas particles.

D. When the pressure of a container of a real gas is less than that predicted by the ideal gas law, this deviation is explained by the fact that the ideal gas law does not correct for the intermolecular forces between the gas particles. Volume corrections, not pressure corrections are made based on the occupied space of the particles. Ideal behavior is seen when the intermolecular forces between particles can be minimized, which occurs at high temperatures and low pressures. This keeps molecules far apart and moving too fast to interact. The shape of the particle has no effect on the deviations of pressure or volume of a real gas. The reduced pressure in a real gas as compared to an ideal gas is due to the attractive forces between the gas particles, which lengthen the path of the particles between collisions with the wall. This reduces the overall number of collisions per unit time, which in turn reduces the measured pressure in the flask.

What is the final phase of a 36 g water vapor sample at 400°C after removal of 15 kJ of energy? [Note: ΔHfus and ΔHvap = 6.0 kJ/mol and 40.6 kJ/mol, respectively; c of solid, liquid, and gaseous water = 38.0 J/mol?K, 74.5 J/mol?K, and 36.0 J/mol?K, respectively] Question 21 Answer Choices A. Aqueous B. Solid C. Liquid D. Gas

D. As we are uncertain of the end phase of the water, let us first calculate the amount of energy required to cool the sample to its boiling point. Given q = mcΔT, we have (2 mol)(36.0 J/mol?K)(300 K) = 21,600 J, or 21.6 kJ, which is far more energy than the amount we removed. Thus, no phase change takes place (choice D is correct). Choice A could have been eliminated before doing any calculations because aqueous is not a phase, but describes a solution in which a solute is surrounded by water.

Which of the following best describes the point indicated in the phase diagram above? Question 28 Answer Choices A. The total quantity of vapor increases over time. B. The total quantity of solid increases over time. C. Equal volumes of solid, liquid, and gas are present. D. Solid, liquid, and gas are in equilibrium.

D. The point indicated on the phase diagram is the triple point, where all three phases indicated are in equilibrium. While the phases are in equilibrium, vapor is likely to occupy the greatest volume (choice C is incorrect). At equilibrium, relative quantities of each phase will remain constant (choices A and B are incorrect).

Table 1: Actual vs. Ideal Pressure for 1.0 mole of steam at 700 K Which of the following are illustrated by the data in Table 1? I. Larger deviations to pressure are seen at higher gas concentrations II. Large deviations occur at standard pressure III. Intermolecular forces have a larger impact on non-ideal behavior in smaller volume Question 1 Answer Choices A. II and III only B. I only C. I, II, and III D. I and III only

Which of the following are illustrated by the data in Table 1? I. Larger deviations to pressure are seen at higher gas concentrations II. Large deviations occur at standard pressure III. Intermolecular forces have a larger impact on non-ideal behavior in smaller volume Items I and III only are illustrated by the data in Table 1. This is a Roman numeral question, so assess each numbered item as true or false, and eliminate answer choices with each decision. Since the number of moles of gas is constant (see table caption), reducing the volume increases the concentration, which shows larger deviations (Item I is true). Standard pressure (1.0 atm) will fall between lines 1 and 2 of the table where minimal deviations are seen, making item II false. Deviations in pressure are due to intermolecular forces (IMFs). Larger concentrations will result in closer particles, which increase the strength of the IMFs and give larger deviations (Item III is true), so the best answer is I and III only.