Chemistry 1A - Spring 2010 - Nitsche - Midterm 3 Solutions

# Chemistry 1A - Spring 2010 - Nitsche - Midterm 3 Solutions...

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Unformatted text preview: Chemistry 1A, Spring 2010 Midterm 3 Solutions Apr 12, 2010 Problems 1. A 1.00 mol sample of CO2 gas is conﬁned in a 20.0 liter container at 25 ◦ C. If 1.00 mol of H2 gas is added, holding the volume and temperature constant, the average kinetic energy of the system will ? A) Increase B) Decrease C) Remain the Same D) Not enough information to answer the question. Kinetic energy depends only on the temperature, so it will remain the same as temperature is held constant. 2. The average kinetic energy of the particles of a gas is directly proportional to A) the rms speed B) the square of the rms speed C) the square root of the rms speed D) the square of the particle mass 3 RT = Ekin = 2 1 mv 2 2 rms ? 3. According to the kinetic-molecular theory, in which of the following gases with the root-mean-square speed of the molecules be the highest at 200◦ C? A) HCl B) Cl2 C) H2 O D) SF6 E) None. The molecules of all gases have the same root-mean-squared speed at any given temperature. Water is the lightest of the choices. Since they will all have the same kinetic energy (as they are at the same ￿ T temperature), the lightest gas must have the highest velocity, as vrms ∝ M 4. According to the kinetic-molecular theory, if the temperature of a gas is raised from 100◦ C to 200◦ C, the average kinetic energy of the gas will ? A) double B) increase by a factor of 1.27 C) increase by a factor of 100 D) decrease by half E) decrease by a factor of 100 1 Kinetic energy depends only on temperature. However, we must use the Kelvin scale, as it is an absolute scale, and we know that kinetic energy cannot be below zero. 100◦ C + 273 = 373K 200◦ C + 273 = 473K Ekin,f Ekin,i 3￿ 2 RTf =￿ 3￿ 2 RT ￿i Tf = Ti ✚ 473✚ K = ✚ 373✚ K = 1.27 Therefore, Ekin,f is 1.27 times larger than Ekin,i 5. Which of the following is NOT part of the kinetic-molecular theory? A) Atoms are neither created nor destroyed by ordinary chemical reactions. B) Attractive and repulsive forces between gas molecules are negligible. C) Gases consist of molecules in continuous, random motion. D) Collisions between gas molecules do not result in the loss of energy. E) The volume occupied by all of the gas molecules in a container is negligible compared to the volume of the container. Kinetic-molecular theory has only to do with gases, not chemical reactions. 6. Which of the following statements is true for ideal gases, but it not always true for real gases? Mark all that apply. A) Collisions between molecules are elastic B) The size of molecules is unimportant compared to the distance between them. C) The volume occupied by the molecules is negligible compared to the volume of the container. D) Average molecular kinetic energy increases with temperature. A,B, and C are all true for ideal gases. However, for real gases, collisions are inelastic (due to intermolecular interactions), the size and volume of the molecules do matter (especially for phase transitions). 7. By what percent does the pressure calculated using the van der Waal’s equation diﬀer from the ideal gas law pressure for a 9.99 mole sample of xenon gas in a 0.828 L container at 496 K? For Xe gas, a = 4.19 L2 atm/mol2 and b = 0.0510 L/mol. The van der Waal’s equation is given below. 2 ￿ PvdW + a n2 V2 ￿ (V − nb) = nRT n2 V2 = = = = nRT (V − nb) nRT n2 −a 2 (V − nb) V ￿ ￿ ✘ 0.08206 L atm ￿496 ✚￿ ￿ ￿ K (9.99✘✘ mol) ✚ L2 atm 9.99 mol 2 mol ✘ K ✘✘✘ − 4.19 0.828 L mol2 0.828 L − 9.99 mol × .0510 L mol ￿ ￿ ✟ ￿ 406.6 ✓ atm L L2 atm mol2 ✟✟ − 4.19￿ ✟ 145.6 ✚ .3185✚ L ￿ mol2 L2 ✟✟ ￿ PvdW + a PvdW PvdW = 1277 atm − 609.9 atm = 667 atm Use the ideal gas law to calculate Pideal Pideal V Pideal = nRT nRT = V = Pideal L atm ✘ ✚ 9.99✘✘ × 0.08206 ✓ ✘✘ × 496✚ mol K mol ✘ ✘K ✚ 0.828✚ L = 491 atm Now, we want to calculate the percent that PvdW diﬀers from Pideal %dif f |Pideal − PvdW | × 100% Pideal |491 atm − 667 atm| = × 100% 491 atm = 35.8% = So, the pressure from the van der Waals equation is A) 35.8% larger than what we would expect from the ideal gas law. The following two questions refer to the phase diagram for water shown below. 3 H2O Solid Liquid P Temperature Gas Tc 8. Which of the following statements is NOT true: A) A Solid to Gas transition is possible by increasing only T. B) A Solid to Liquid transition is possible by increasing only T. C) A Liquid to Gas transition is possible by increasing only T. D) A Solid to Liquid transition is possible by increasing only P. E) A Solid to Gas transition is possible by increasing only P. The only way to go from Solid to Gas is by increasing temperature or decreasing pressure. 9. The critical point (above which there is no gas/liquid phase transition) of water occurs at 375 ◦ C and 218.0 atm. the liquid vaporizes to gas at 1.00 atm and 100 ◦ C At what pressure are water liquid and vapor at equilibrium at 250 ◦ C? D) 185.00 atm. If we follow the liquid/gas equilibrium line for water, we see that as temperature decrease, so does the pressure for liquid/gas equilibrium. Therefore, at 250 ◦ C, the pressure for equilibrium must be less that that of the critical point, but greater than that of 100 ◦ C. 185.00 atm as the only answer which is between 1.00 atm and 218.00 atm of pressure. 10. 2.0 moles of a monatomic ideal gas are compressed in a cylinder at a constant pressure of 10 atm from 10 L to 5.0 L. What is the work done on the gas for this process, in kJ? w = −Pe xt∆V w = −Pe xt (Vf − Vi ) = − (10 atm) (5.0 L − 10 L) 100￿ J 1 kJ ￿ ✘ = 50✘✘✘ × L atm = 5.0 kJ ✘× 1✘✘✘ L atm 1000￿ J ￿ D) 5.0 kJ of work is done on the gas. The sign of the work should be positive, since for compression, positive work is done on the system, and negative work is done by the surroundings. 4 11. The internal energy of a system is always increased by ? A) adding heat to the system B) having the system do work on the surroundings C) withdrawing heat from the system D) adding heat to the system and having the system do work on the surroundings E) a volume compression 12. Which one of the following is an exothermic process? A) ice melting B) water evaporating C) boiling soup D) condensation of water vapor E) Ammonium thiocyanate and barium hydroxide are mixed at 25 ◦ C: the temperature of the surrounding water drops. 13. Which of the following is a statement of the ﬁrst law of thermodynamics? 2 A) Ek = mv 2 B) A negative ∆H corresponds to an exothermic process C) ∆E = Ef inal − Einitial D) Energy lost by the system must be gained by the surroundings. E) 1 cal = 4.184 J 14. Propanol (pro) burns cooler than isopropanol (iso). What can you conclude about the isomerization of propanol to isopropanol? propanol isopropanol B) ∆H ◦ > 0 ◦ Propanol burns cooler than isopropanol, which means that isopropanol has a higher ∆Hf than propanol. ◦ would have to be positive. Therefore, to go from propanol to isopropanol, ∆H 15. The value of ∆H ◦ for the reaction below is -790 kJ. The enthalpy change accompanying the reaction of 0.95 g of S is kJ. 2 S(s) + 3 O2(g) → 2 SO3(g) 0.95✟✟ × gS 1✘✘✘ mol ✘ −790 kJ S × = C) − 12 kJ ✟ 32✟ S g 2✘✘✘ mol ✘ S 5 16. ∆ H for the reaction: C6 H4 (OH)2 (aq) + H2 O2 (aq) → C6 H4 O2 (aq) + 2 H2 O(l) is kJ, given the data below. → C6 H4 (OH)2 (aq) ∆H = -177.4 kJ ∆H = 191.2 kJ ∆H = 241.8 kJ ∆H =-43.8 kJ → H2(g) + O2(g) → H2(g) + 1 O2(g) 2 → H2 O(l) C6 H4 O2 (aq) + H2(g) H2 O2 (aq) H2 O(g) H2 O(g) To get the correct equation, reverse the ﬁrst reaction, multiply the third reaction by 2 and reverse, and double the fourth reaction, as follows: C6 H4 (OH)2 (aq) 2 H2(g) + O2(g) 2 H2 O(g) H2 O2 (aq) → C6 H4 O2 (aq) + H2(g) → H2(g) + O2(g) → 2 H2 O(g) → 2 H2 O(l) ∆H = 177.4 kJ ∆H = 191.2 kJ ∆H = -483.6 kJ ∆H = -87.6 kJ So, when we add these together, we get C) ∆ H = -202.6 kJ 17. Which one of the following statements is true? A) Enthalpy is an intensive property. B) The enthalpy change for a reaction is independent of the state of the reactants and products. C) Enthalpy is a state function. D) H is the value of q measured under conditions of constant volume. E) The enthalpy change of a reaction is the reciprocal of the ?H of the reverse reaction. 18. A 46.2 g sample of copper is heated to 95.4 ◦ C and placed in a calorimeter containing 75.0 g of water at 19.6 ◦ C. The ﬁnal temperature of the water is 21.8 ◦ C. Assuming the calorimeter does not absorb any heat, what is the heat capacity of copper, in g J ? The heat capacity of water is 4.184 g J . K K since the calorimeter is ideal, we have: heat lost by copper = heat gained by water −mCu CCu ∆TCu = mW CW ∆TW mW CW ∆TW CCu = −mCu ∆TCu = = −qCu = qW 75.0 g water × 4.184 J × (21.8◦ C − 19.6◦ C) g ◦C −46.2 g Cu × (21.8◦ C − 95.4◦ C) 75.0✚ × 4.184 J✟ × (2.2◦￿ g C) ✚ g ◦C ✟✟ −46.2 g × (−73.6◦ C) 6 = 690 J 3400 g◦ C CCu = B) 0.203 J g◦ C 19. The ∆H for the solution process when solid sodium hydroxide dissolves in water is -44.4 kJ/mol. When a 13.9 g sample of NaOH dissolves in 250.0 g of water in a coﬀee-cup calorimeter, the temperature increases ◦ C. Assume that the solution has the same speciﬁc heat as liquid water, i.e., J . from 23.0 ◦ C to gK (*note: on the exam, the ∆H was listed as +44.4 J/mol. This would have caused the water to decrease in temperature. Therefore, everyone was given credit for this problem) heat given oﬀ by reaction = heat absorbed by water solution (since water increases in temperature) −qrxn = qsol ￿ ￿ ✭ ✭✭ ✟ mol NaOH −44.4✟ kJ 1000 J ✘ 1✭✭✭ −qrxn = − 13.9✘✘✘✘ × g NaOH × × = 15430 J ✭ ✘ ✭ ✟ 40✘✘✘✘ g NaOH 1✭✭✭✭ mol NaOH 1✟ kJ 15430 J = qsol 15430 J = msol Csol ∆Tsol 15430 J = (250.0 g water + 13.9 g NaOH) × 4.18 ✘ 15430 J = 263.9✘✘ × 4.18 g sol J × (Tf − (23.0 + 273) K) gK J × (Tf − 296 K) K J 15430 J = 1103 × Tf − 326500 J K J 1103 × Tf = 311070 J K Tf = 310 K − 273 = D) 37.0◦ C 15430 J = 1103 J × (Tf − 296 K) gK ✁ 20. The combustion of titanium with oxygen produces titanium dioxide: Ti(s) + O2(g) → TiO2(s) When 2.060 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 ◦ C to 91.60 ◦ C. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kJ/K. The heat of reaction for the combustion of Ti in this calorimeter is kJ/mol. heat given oﬀ by combustion = heat gained by water + heat gained by calorimeter −qrxn = qcal 7 −qrxn qrxn = Kcal ∆Tcal kJ = 9.84 ◦ (91.60◦ C − 25.00◦ C) C kJ = 9.84 ◦ (66.60◦ C) C kJ = 9.84 ◦ (66.60◦￿ C) C ￿ = −655.3 kJ ✟ 2.060✟✟ × g Ti 1 mol Ti ✟ = 0.04303 mol Ti 47.87✟✟ g Ti −655.3 kJ kJ = qrxn = E) − 15200 0.04303 mol Ti mol 21. The following process of bond-breaking → bond formation corresponds to the combustion of which compound? 3 C-H + O-H + C-O + A) Ethanol B) Ethane 3 2 O=O → 2 C=O + 4 O-H D) Methane E) Formaldehyde C) Methanol C) Methanol The combustion of methanol has the following equation: CH3 OH + 3 O2 → CO2 + 2 H2 O 2 Which matches the bonds broken and bonds formed from the question 22. Methane can react with a ﬂuorine atom to produce a methyl radical and hydrogen ﬂuoride. H°rxn = -153 kJ/mol 8 Based on the ∆H◦ , which of the following is true? rxn A) Breaking chemical bonds releases energy. B) All reactions involving bond-breaking and bond formation are exothermic. C) The H-F bond is stronger than the C-H bond. D) The H-F bond is weaker than the C-H bond. E) The heat of the reactants is absorbed by the products. 23. Three balloons were ﬁlled with the following mixtures. Which balloons looses the most heat when the contents undergo combustion reactions? A) 2 moles C8 H18 B) 3 moles C8 H18 C) 2 moles C8 H18 and 2 moles O2 B) has the most moles of hydrocarbon, so it will have the most heat released upon combustion. 24. Which process is accompanied by the largest increase in entropy? I II III IV V VI A) I → II B) I → III C) II → III D) III → IV E) V → VI I has the most order, so the largest change in entropy would be from I to III, as III has the most disorder. 25. Which of the following processes have a ∆S > 0 A) CH3 OH(l) → CH3 OH(s) B) N2 (g) + 3 H2 (g) → 2 NH3 (g) C) CH4 (g) + H2 O(g) → CO(g) + 3 H2 (g) D) Na2 CO3 (s) + H2 O(g) + CO2 (g) → 2 NaHCO3 (s) E) All of the above processes have a ∆S > 0. C) is the only reaction where the number of moles of gas increase, so it is the only one with a positive ∆S 26. Calculate ∆Srxn (J/K) for the reaction of 2.30 moles of NH3 at standard conditions. The S◦ for each species is shown below the reaction. 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2 O (g) 9 S◦ ￿ J mol K ￿ 192.5 205.1 210.8 188.8 ◦ ∆Srxn = ◦ ∆Srxn ◦ ◦ ◦ ◦ = 4SNO + 6S H O − 4SNH − 5S O 2 (g) (g) 3 (g) 2 (g) ￿ ￿ ￿ ￿ ￿ ￿ ￿ ￿ J 188.8J J J = 4 210.8 +6 − 4 192.5 − 5 205.1 mol K mol K mol K mol K J J J J = 843.2 + 1132.8 − 770 − 1025.5 mol K mol K mol K mol K J = 180.5 mol K ￿ ◦ ∆Sproducts − ￿ ◦ ∆Sreactants So, for 2.30 moles of NH3 180.5 J mol K = C) 103.8 J ✘✘✘ × 2.30✘✘ NH3 mol ✘ mol K 4✘✘✘✘ 3 mol NH 27. Above what temperature does the following reaction become non spontaneous? 2 H2 S(g) + 3 O2 (g) 2 SO2 (g) + 2 H2 O (g) ∆ H = -1036 kJ; ∆ S = -153.2 J/K ∆G = ∆ H − T ∆S 0 = ∆ H − T ∆S ∆H T= ∆S −1036 kJ × 1000 J 1 kJ = J −153.2 k T = 6762 K This is the temperature at which ∆G = 0. When the temperature is above this, the T ∆S term, which is positive, will dominate. This will make the reaction non spontaneous (∆G > 0) above A) 6.762 × 103 K. 28. Calculate ∆Grxn (kJ) at 298 K under the conditions shown below for the following reaction. SO3 (g) + H2 O(g) → H2 SO4 (l) 10 ∆G◦ = −90.5 kJ at 298 K PSO3 = 0.20 atm, PH2 O = 0.88 atm ∆G = ∆G◦ + RT ln Q J 1 kJ 1 ✁ ✚ × × 298✚ × ln K PSO3 PH2 O mol ￿ 1000￿ K J ￿ kJ 1 = −90.5 kJ + 2.48 × ln mol (0.20 atm) (0.88 atm) = −90.5 kJ + 2.48 kJ × ln 5.682 = −90.5 kJ + 8.3145 ∆G = = −90.5 kJ + 4.30 kJ D) − 86.2 kJ 29. The reaction between carbon tetrachloride and water to form carbon dioxide and hydrogen chloride has a ∆G◦ value of -232 kJ/mol, and so it is thermodynamically favored. But when you mix carbon tetrachloride with water, no change is observed. Which one of the following can best explain this? A) The reaction is not favored thermodynamically. B) It is a slow reaction kinetically. C) The activation energy of the reaction is too small. D) The reaction is not spontaneous. 30. At equilibrium, . A) all chemical reactions have ceased B) the rates of the forward and the reverse reactions are equal C) the rate constants of the forward and reverse reactions are equal D) the value of the equilibrium constant is 1 E) the limiting reagent has been consumed 31. Consider the following reaction at equilibrium: 2 CO2 (g) ￿ 2 CO (g) + O2 (g) ∆H = -514 kJ . Le Chatelier’s principle predicts that an increase in temperature will A) increase the partial pressure of O2(g) B) decrease the partial pressure of CO2(g) C) decrease the value of the equilibrium constant D) increase the value of the equilibrium constant E) increase the partial pressure of CO The reaction is exothermic, so heat can be treated as a product. Therefore, increasing the temperature will cause the reaction to shift towards the reactants, decreasing the value of the equilibrium constant (as changing temperature will change the equilibrium constant). 11 32. Consider the following reaction at equilibrium: C (s) + H2 O (g) ￿ CO (g) + H2 (g) Which of the following conditions will increase the partial pressure of CO? A) decrease the partial pressure of H2 O(g) B) removing H2 O(g) from the system C) decreasing the volume of the reaction vessel D) decreasing the pressure in the reaction vessel E) increasing the amount of carbon in the system To see this eﬀect, consider the eﬀect of halving the pressure of all gases on Q K= Q= = = PCO PH2 PH2 O 1 1 2 PCO × 2 PH2 1 2 PH2 O 1 4 PCO PH2 1 2 PH2 O 1 PCO PH2 2 PH2 O 1 Q= K 2 Therefore, since Q < K, the equilibrium must shift towards the products to re-establish equilibrium, which will increase PCO . 33. The eﬀect of a catalyst on an equilibrium is to . A) increase the rate of the forward reaction only B) increasing the equilibrium constant so that products are favored C) slow the reverse reaction only D) increase the rate at which equilibrium is achieved without changing the composition of the equilibrium mixture E) shift the equilibrium to the right A catalyst will make the reactions occur more quickly, without aﬀecting the thermodynamics. 34. Choose the statement below that is true. A) If K > 1, ∆G◦ is positive. rxn B) If K < 1, ∆G◦ is negative. rxn C) If ∆G◦ = 0, the reaction is at equilibrium. rxn D) If ∆Grxn = 0, the reaction is at equilibrium. E) None of the above statements are true. 12 35. Use the free energy of formation given below to calculate the equilibrium constant K for the following reaction at 298 K. 2 HNO3 (aq) + NO (g) ￿ 3 NO2(g) + H2 O(l) ∆G◦ (kJ/mol) -110.9 f 87.6 51.3 -237.1 ∆G◦ rxn = = = = ∆G◦ rxn = ￿ ∆G◦ products − ∆G◦ reactants f, f, ◦ ◦ 3∆Gf, NO + ∆Gf, H O − 2∆G◦ HNO − ∆G◦ NO f, f, 2 3 ￿2 ￿ ￿ ￿ ￿ ￿ ￿ ￿ kJ kJ ✘ ✘ ✘ ✘ ✘ 51.3 ✘ −237.1 ✘ −110.9 kJ − 1✘✘ 87.6 kJ 3✘ mol + 1✘ mol − 2✘ mol mol mol mol mol mol ✟✟ ✟✟ ✟✟ ✟✟ 153.9 kJ − 237.1 kJ + 221.8 kJ − 87.6 kJ ✟ 1000 J = 51000 J 51.0✟ × kJ ✚ 1✚ kJ ∆ G◦ rxn = −RT ln Keq ∆ G◦ rxn Keq = exp − RT = exp − Keq = ￿ = e−20.58 51000 J 8.3145 J × 298 K mol K C) 1.15 × 10−9 36. Consider the following equilibria and equilibrium constants: X￿Y Y￿Z What is the equilibrium constant, K, for X￿Z? K1 = 5.0 K2 = 5.0 K1 = K2 = K= = = [Y ] = 5.0 [X ] [Z ] = 5.0 [Y ] [Z ] [X ] [Z ] [Y ] [Y ] [X ] K1 × K2 = E) 25.0 13 37. At 22◦ C, Kp = 0.070 for the equilibrium: NH4 HS (s) ￿ NH3 (g) + H2 S (g) A sample of solid NH4 HS is placed in a closed vessel and allowed to equilibrate. Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH4 HS remains. 0.070 = PN H3 · PH2 S Now, for every 1 mol of NH4 HS that reacts, there is 1 mole of NH3 and 1 mole of H2 S produced. Therefore, for every x moles of NH3 , there are x moles of H2 S 0.070 = PN H3 · PH2 S 0.070 = x2 √ x= 0.070 Kp = PN H3 · PH2 S x = 0.26 atm So, ammonia will have a partial pressure of A) 0.26 atm. 38. In the experiment ”‘How Hot is that Flame”’, 100.0 g of room temperature water (25.0◦ C) was put in a calorimeter. 60.0 g of cold water (3.0◦ C) was added. When the temperature of the entire setup is constant, it was 17.5◦ C. The speciﬁc heat of water is 4.184 g JC . What is the heat capacity of the calorimeter, Kcal ? ◦ heat lost by room temperature water + heat lost by calorimeter = heat gained by cold water −qRT − qcal = qCW qcal = −qCW − qRT mcal Ccal ∆Tcal = −qCW − qRT Kcal = mcal Ccal since the mass and heat capacity of the calorimeter are both constants Kcal ∆Tcal = −qCW − qRT −qCW − qRT Kcal = ∆Tcal −mCW Cw ∆TCW − mRT Cw ∆TRT Kcal = ∆Tcal ￿ ￿ ￿ ￿ − (60.0✚ 4.184 J g) (17.5◦ C − 3.0◦ C) − (100.0✚ 4.184 J g) (17.5◦ C − 25.0◦ C) ✚ ✚ g ◦C g ◦C ✁ ✁ Kcal = (17.5◦ C − 25.0◦ C) ￿ ￿ ￿ ￿ − 251.0 ◦J (14.5◦￿ − 418.4 ◦J (−7.5◦￿ C) C) ✚ ✚ C C ✚ ✚ Kcal = (−7.5◦ C) 14 Kcal = Kcal Kcal −3639.5 J + 3138.0 J (−7.5◦ C) −501.5 J = (−7.5◦ C) J = B) 67 ◦ C 39. A student explored the thermodynamic properties of the dissolution of borax using titrations at diﬀerent J temperatures. The following plot was obtained. What is ∆S ◦ in for the dissolution of borax? mol K Van't Hoff Plot for the Dissolution of Borax y = -16870x + 53.727 2.00 1.00 0.00 3.10E-03 -1.00 lnKsp -2.00 -3.00 -4.00 -5.00 -6.00 -7.00 R2 = 0.9978 3.20E-03 3.30E-03 3.40E-03 3.50E-03 3.60E-03 1/T We can use the Van’t Hoﬀ equation: ∆H ◦ ln K = − R which shows that the y intercept will be ￿￿ 1 ∆S ◦ + T R ∆S ◦ 1 for a plot of ln K vs. . Therefore: R T ∆S ◦ = 53.727 R ◦ ∆S = 53.727 · R ∆S ◦ = ∆S ◦ = 53.727 · 8.314 J mol K J C) 446.69 mol K 40. A student has a solution of potassium chromate (K2 CrO4 ) in a test tube. The solution is yellow. To this test tube, several drops of a solution of sulfuric acid are added. Given the reactions below, predict what changes will occur after the addition of acid and why the changes occur. 15 K2 CrO4 (s) ￿ 2 K+ + CrO−2 (yellow) (aq) 4 (aq) H2 SO4 (aq) ￿ H+ + HSO−2 (aq) 4 (aq) 2 H+ + 2 CrO−2 (yellow) ￿ Cr2 O−2 (orange) + H2 O (￿) (aq) 4 (aq) 7 (aq) A) The solution remains yellow because sulfuric acid is colorless B) The solution remains yellow because no additional reactions take place C) The solution turns a lighter yellow because it is more dilute after adding acid D) The solution turns orange because the acid reacts with the chromate to convert it to dichromate which is orange. 16 ...
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