CHEM 1331 Chp 6, 7, 8

Which of the following compounds is soluble in water? A.BaSO4 B.Cu2O C.AgBr D.NaClO3 E.Hg2Cl2

D. NaClO3

What is the change in internal energy (in kJ) of a system that absorbs 6.457 kJ of heat from its surroundings and has 2.638 kcal of work done on it?

17.49 ± 2% ΔE=q + w for work being done ON it. Convert work to kJ. 2.638 kcal x (4.184 kJ/1 kcal)= 11.04 kJ of work Plug into the formula. 6.457 kJ + 11.04 kJ= 17.5

How many moles of H+ ions are present in the following aqueous solution?363 mL of 6.95 M hydriodic acid

2.52 ± 2% *Recall Molarity= moles/L so mole=M x L (6.95 M) x (.363)= 2.52 mol

If 5.33 x 10-3 mol of argon occupies a 543-mL container at 179.1°C, what is the pressure (in torr)?

276.9 ± 2% P=nRT (5.33 x 10^-3 mol)(0.082 atmxL/molxK)(452K)/(.543 L)=.364 atm 1 atm=760 torr *** .364 x (760 torr/ 1 atm)= 276.6 torr

A sample of gas is in a 50.0-mL container at a pressure of 645 torr and a temperature of 25°C. The entire sample is heated to a temperature of 35°C and transferred to a new container whose volume is 96.5 mL. The pressure of the gas in the second container is about: A.1.29 × 103 torr B.345 torr C.323 torr D.468 torr E.72 torr

413 ± 2%

The heat capacity of a bomb calorimeter is found to be 5.68 kJ C^-1. When a 1.652 g sample of glucose, C6H12O6 , is burned in the calorimeter, the temperature of the calorimeter increases from 24.56 C to 29.10 C. Calculate the H for the following reaction: C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(g) A.-2.81 x kJ B.-25.8 kJ C.15.6 kJ D.-42.6 kJ E.25.8 kJ

A. -2.81 x kJ

The heat capacity of a bomb calorimeter is found to be 5.68 kJ . When a 1.652 g sample of glucose, , is burned in the calorimeter, the temperature of the calorimeter increases from 24.56 C to 29.10 C. Calculate the H for the following reaction: A.-2.81 x kJ B.-25.8 kJ C.15.6 kJ D.-42.6 kJ E.25.8 kJ

A. -2.81 x kJ

Calculate the energy change in kJ/mol for the reaction Li(g) + F(g) => Li+(g) + F-(g) (separated ions) using the following information: Li(g) → Li+(g) + e- +520 kJ/mol F(g) + e- → F-(g) -328 kJ/mol

192

One step in the synthesis of nitric acid is the conversion of ammonia to nitric oxide. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) Calculate ΔH°rxn for this reaction. ΔH°f [NH3(g)] = -45.9 kJ/mol; ΔH°f [NO(g)] = 90.3 kJ/mol; ΔH°f [H2O(g)] = -241.8 kJ/mol. A.-906.0 kJ B.-197.4 kJ C.-105.6 kJ D.197.4 kJ E.906.0 kJ

A. -906.0 kJ ΔH°rxn = ΔH°products - ΔH°reactants ΔH°rxn = ((4*90.3) + 6*-(241.8)) - 4*(-45.9) = -906 kJ

Perform the indicated conversion: 31.95 kJ = _______________ kcal A.7.636 kcal B.13.37 x 10-2 kcal C.0.1310 kcal D.7.636 x 103 kcal E.1.337 x 102 kcal

A. 7.636 kcal Convert kJ > kcal *1 kJ=1000J *4.184 J=1 cal *1000 cal=1 kcal (31.95 kJ) x (1000 J/1 kJ) x (1 cal/ 4.184 J) x (1kcal/1000 cal)= 7.636 kcal

Consider the following gas phase reaction: 2NO(g) + O2(g) --> 2NO2(g) 400. mL of NO at STP is reacted with 500. mL of O2 at STP. Calculate the volume of the reaction mixture at STP after the reaction goes to completion. A.700 mL B.800 mL C.900 mL D.1300 mL E.100 mL

A. 700 mL

A gauge on a tank of compressed helium reads 1850 psi (pounds per square inch). What is this pressure in mm Hg? A.9.57 x 10^4 B.1.41 x 10^6 C.126 D.35.8 E.2.72

A. 9.57 x 10^4 mmHg=psi value x 51.749 1850 x 51.749= 957 x 10^4

Data: SO2(g) + 1/2O2(g) => SO3(g) ΔH = -99.1 kJ. Given the above data, calculate the enthalpy change ΔH when 89.6 g of SO2 is converted to SO3. A.-69.3 kJ B.-139 kJ C.69.3 kJ D.139 kJ E.-111 kJ

B. -139 kJ grams > mols> kJ where 1 mol of SO2 undergoing combination gives off= -99.1kJ of heat. (89.6g SO2) x (1 mol/64 g) x (-99.1 kJ/1 mol)= -139 kJ

Calculate ΔHrxn for the reaction below given: ΔHf [AsH3(g)] = 66.4 kJ/mol; ΔHf [H3AsO4(aq)] = -904.6 kJ/mol; ΔHf [H2O(l)] = -285.8 kJ/mol; H3AsO4(aq) + 4H2(g) => AsH3(g) + 4H2O(l) A.-685.2 kJ B.-172.2 kJ C.172.2 kJ D.685.2 kJ E.-1123 kJ

B. -172.2 kJ ΔHf Sum of Products - Reactants 66.4 + 4(-285.8) - (-904.6)= -1076.8 + 904.6 = -172.2 kJ

Calculate the energy change (change in enthalpy) for the decomposition of hydrogen peroxide: 2H2O2 -> 2H2O + O2 Given these bond energies. H-H: 436 O-H: 463 O-O: 146 O=O: 498 A.-314 kJ B.-206 kJ C.-498 kJ D.-255 kJ E.+314 kJ

B. -206 kJ *H2O2 has a single bond between the Oxygens. 2(146) + 2(2 x 463) - 2(2 x 463) + 498 = 2144 - 2350 = -206 kJ

2Fe(s) + O2(g) => 2FeO(s) ΔH = -544.0 kJ 4Fe(s) + 3O2(g) => 2Fe2O3(s) ΔH = -1648.4 kJ Fe3O4(s) => 3Fe(s) + 2O2(g) ΔH = +1118.4 kJ Given the data above, determine the heat of reaction, ΔH, for the reaction below: Reaction: Fe2O3(s) + FeO(s) => Fe3O4(s) A.-1074.0 kJ B.-22.2 kJ C.+22.2 kJ D.+249.8 kJ E.+1074.0 kJ

B. -22.2 kJ (1/2)(1648.4) + (1/2)(544) + (-1118.4) 824 + 272 + (-1118.4)= -22.4 kJ

Ethylene glycol, used as a coolant in automotive engines, has a specific heat capacity of 2.42 J/(g-C). Calculate q when 3.65 kg of ethylene glycol is cooled from 132°C to 85°C. A.-1900 kJ B.-420 kJ C.-99 kJ D.-0.42 kJ E.-4.2 × 10¯6 kJ

B. -420 kJ q= mcΔT 1. Check units: 3.65 kg x (1000g/1 kg)= 3650 g 2. Plug in (3650) (2.42) (85-132)= -4.15 x 10^5 J 3. Convert to kJ -4.15 x 10^5 J x (1 kJ/1000 J)= -415 kJ

How many moles of KClO3 at STP will be needed to produce 22.4 L of O2 gas according to the following equation: 2KClO3(s) => 2KCl(s) + 3O2(g) A.0.33 B.0.67 C.1.0 D.2.0 E.3.0

B. 0.67 ** 1 mol= 22.7 L We can use that to convert from L -> mols 22.4 L x (1 mol x 22.7 L) x ( 2 mol KClO3 / 3 mol O2)= 0.65 mols KClO3

Exactly 248.0 J will raise the temperature of 10.0 g of a metal from 25.0°C to 60.0°C. What is the specific heat capacity of the metal? A.1.41 J/g°C B.0.709 J/g°C C.12.4 J/g°C D.59.3 J/g°C E.none of these

B. 0.709 J/g°C q=mcΔT 248 J= 10 x c x (60-25)= solve for c. 248/ 350=c=0.709

What is the effect of the following on the volume of 1 mol of an ideal gas?The pressure is reduced by a factor of four (at constant T). A.V decreases by 75% B.V doubles C.V increases 16 fold D.V does not change since n and T are constant E.V increases 4 fold

E. V increases 4 fold Volume and pressure are inversely proportional, so an increase in pressure decreases volume. Here There is less pressure, so volume will increase.

Convert 380 torr to kilopascals. A.50.7 kPa B.2.64 kPa C.38.5 kPa D.85.4 kPa E.5.07 x 104 kPa

A. 50.7 kPa 380 torr x (1 atm/760 torr) x (101.3 kPa/1 atm)= 50.7 kPa

Use the given average bond dissociation energies to estimate ΔE (kJ) for the reaction of methane, CH4(g), with fluorine according to the equation below. Enter your answer as the nearest whole number without units. CH4(g) + 2 F2(g) → CF4(g) + 2 H2(g) Bond, Kj/mol C-F: 450 C-H: 410 F-F: 158 H-H: 436

-716 Bond broken- bonds formed (4 x 410) + (2 x 158) - (4 x 450) + (2 x 436)= 1956-2672= -716

Use the given average bond dissociation energies to estimate ΔE (kJ) for the reaction of methane, CH4(g), with fluorine according to the equation below. Enter your answer as the nearest whole number without units. CH4 + 2 F2 -> CF4 + 2 H2 Bond; kJ/mol C-F; 450 C-H; 410 F-F; 158 H-H; 436

-716 Bonds broken-Bonds formed= 1956-2672= -716

Thermal decomposition of rail car load of limestone to lime and carbon dioxide requires 4.14 x 106 kJ of heat. Convert this energy to calories. Once you have the value, determine its log (base 10) and enter this in decimal format with three decimal places. Do not include units.

8.995 ± 0.05

A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715, what is the mass percent of Ar?

86.2% 1. Find total # of mols with n=PV/RT n=(1.040 x 10)/(0.082 x 318)= 0.40 mols total Determine # of mols of Ar by using the mole fraction formula. 0.715= (x mols of Ar/0.40 mols)= 0.29 mols 2. Convert mols to grams 0.29 molsof Ar x (40g/ 1 mol)= 11.6 g Ar 3. What is the total mass though? We still need grams of CH4. We can find mols of CH4 by subtracting from 0.40 mols. 0.40 mols - 0.29 mol Ar= 0.11 mol CH4 Convert to grams> 0.11 mol x (16g/1 mol)= 1.76 g CH4 4. Total mass= 1.76g + 11.6g= 13.36g Now we can solve for mass percent! 5. 11.6g Ar/ 13.36g= 0.87 x 100= 87%

Two aircraft rivets, one of iron and the other of copper, are placed in a calorimeter. The data for the metals are as follows: Iron: Mass= 30.0; Initial T (°C);Initial T (°C)= 0.0; c (J/g·K)=0.450 Copper: Mass= 20.0=;Initial T (°C);Initial T (°C)= 100.0;c (J/g·K)=0.387 Mass (g) Assuming that the calorimeter absorbs no heat, which of the following statements is TRUE? A.Heat flows from Cu to Fe and qCu = -491J B.The final temperature is 63.5 °C C.Heat flows from Fe to Cu D.qFe = -491J E.None of these is correct

A. Heat flows from Cu to Fe and qCu = -491J a) Copper is at a higher temperature, so the flow of heat will take place from copper to iron. Heat is a form of energy, which always flows from higher temperature to lower temperature. b) To determine the actual final temperature, the heat capacity of the calorimeter must be known. A calorimeter constant refers to a constant, which quantifies the heat capacity of a calorimeter. It may be determined by using a known amount of heat to the calorimeter and measuring the corresponding change in temperature of the calorimeter.

When 1 mol of KBr(s) decomposes to its elements, 394 kJ of heat is absorbed.Which choice is a balanced thermochemical equation for this reaction? A.KBr(s) => K(s) + 1/2Br2(l) ΔHrxn = 394 kJ B.KBr(s) => K(s) + 1/2Br2(l) ΔHrxn = -394 kJ C.2KBr(s) => 2K(s) + Br2(l) ΔHrxn = 394 kJ D.2KBr(s) => 2K(s) + Br2(l) ΔHrxn = -394 kJ E.None of these

A. KBr(s) => K(s) + 1/2Br2(l) ΔHrxn = 394 kJ

The following elements are grouped with either a maximum or a minimum oxidation state and their redox function. Which combination is incorrect? A.N, +5, reducing agent B.S, -2, reducing agent C.Cl, +7, oxidizing agent D.Se, +6, oxidizing agent E.Br, -1, reducing agent

A. N, +5, reducing agent

Molecular size in real gases causes positive deviations from the PV/RT ratio of an ideal gas. A.True B.False

A. True

The kinetic-molecular theory provides a model for gases at the microscopic level that explains the physical properties at the macroscopic level. A.True B.False

A. True

Consider two 1-L samples of gas, one H2 and one O2, both at 1 atm and 25°C. Compare the samples in terms of the characteristics listed. Time for a given fraction of O2 molecules to effuse ___ time for a given fraction of H2 molecules to effuse A.is greater than B.is less than C.is equal to D.is unrelated to E.None of these

A. is greater than As the mass increases=decrease in rate.

Which species is the oxidizing agent in the following equation? K + O2 ---> KO2 A.K B.O2 C.KO2 D.This is not an oxidation-reduction reaction

B. O2 Determine oxidation for each one. K + O2 ---> KO2 K= monoatomic ions= oxidation #= charge, so its +1 O2=diatomic/atoms bonded to the same atom= 0 KO2; where O2=-4 so K will be +4= K is going from +1 > +4 its being oxidized O2 is going from 0 > -4 its being reduced. Oxidizing agent- "which one is being reduced?"> O2

For the process described, compare Δ H with Δ E of the system.A solid yields a mixture of gases in an exothermic reaction that takes place in a container of variable volume. A.ΔH is less than (more negative) Δ E of the system. B.ΔH is greater than ΔE of the system. C.Δ H is equal to ΔE of the system. D.can't be determined without more information

B. ΔH is greater than ΔE of the system.

What is the oxidation number of chlorine in ClO4- ? A.+ 8 B.- 8 C.+ 7 D.+ 5 E.+ 9

C. + 7 ClO4 - O has a oxidation of -2, so -2 x 4 molecules of Oxygen= -8 total charge on Oxygen. Now we can solve for Cl. Where x= Cl oxidation #. So x + -8= -1 total charge. Solve for x x= -1 +8 x=+7

What is the average oxidation number of the irons in Fe3O7? Note that average oxidation numbers are not restricted to integer values. A.+3 B.+7/3 C.+4 (2/3) D.+5 E.+6/7

C. +42/3 Recall that Oxygen has a -2 oxidation #, -2 x 7 oxygen atoms= -14 charge that needs to balanced out by the 3 Fe atoms. So we can do 14/3= + 4.6= + 14 (2/3) to balance out the -14 charged Oxygen atoms to have an overall neutral charge of 0.

Estimate the standard energy change ΔE for the reaction below: C2H4(g) + H2(g) => C2H6(g) Data: Bond energies in kJ/mol: H-H, 432; C-H, 415; C-C, 345; C=C, 615 A.-280 kJ B.-180 kJ C.-128 kJ D.+974 kJ E.-228 kJ

C. -128 kJ Bonds broken- bonds made 615 + 4(415) + 432 - 345 + 6(415) = -128

Add the following thermochemical equations: (1) P4(s) + 6Cl2(g) => 4PCl3(g) ΔH = -1148 kJ (2) 4PCl3(g) + 4Cl2(g) => 4PCl5(g) ΔH =-460 kJ (3) Overall equation = ? ΔH overall = ? Select the correct ΔH for the overall process (equation 3) A.-688 kJ B.688 kJ C.-1608 kJ D.1608 kJ E.None of these is within 5% of the correct answer

C. -1608 kJ summing both equation gives P4(s) + 6Cl2(g) +4PCl3(g) + 4Cl2(g) => 4PCl3(g) + 4PCl5(g) but 4PCl3(g) is in both sides , so is cancelled out P4(s) + 6Cl2(g)+ 4Cl2(g) => 4PCl5(g) summing the Cl2(g) P4(s) + 10Cl2(g) => 4PCl5(g) → overall reaction since we sum both equations the ΔH of the overall reaction is also the sum of the ΔH of the individual reactions. ΔH overall = ΔH1 + ΔH2 = (-1148 kJ) + (-460 kJ) = (-1608 KJ)

The pressure of sulfur dioxide in a container is 159 kPa. What is this pressure in atmospheres? A.0.209 atm B.0.637 atm C.1.57 atm D.21.2 atm E.15900 atm

C. 1.57 atm 1 atm= 101, 325 Pa= 101.3 kPa 159 kPa x (1 atm/101.3 kPa)= 1.57 atm

What is the ratio of effusion rates for O2 to Kr? A.2.619 B.0.6179 C.1.618 D.0.3819 E.None of these is within 5% of the correct answer

C. 1.618

Calculate the work associated with the compression of a gas from 121.0 L to 80.0 L at a constant pressure of 16.7 atm. A.2.46 L atm B.-2.46 L atm C.685 L atm D.-685 L atm E.101 L atm

C. 685 L atm w=-PΔV w=-(16.7)(80-121)= (16.7)(-41)=685 L x atm

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 torr. What are the partial pressures of each gas? A.Phalothane = 778 torr, PO2 = 84 torr B.Phalothane = 162 torr, PO2 = 700 torr C.Phalothane = 84 torr, PO2 = 778 torr D.Phalothane = 155 torr, PO2 = 707 torr E.Phalothane = 707 torr, PO2 = 155 torr

C. Phalothane = 84 torr, PO2 = 778 torr *I rounded up some numbers so they're off a little but overall correct. 1. Formulas: PA= XA + PT So the partial pressure= mole fraction of the element + Total Pressure Mole fraction (XA) can be found by XA= (nA/nT) or (PA/PT) We can use either the moles/pressure of the target element divided by the total amount of moles/pressure. 1. Find the mols of each by converting Grams > mols (15g) x (1 mol/ 197.4 g)= 0.076 mol (22.6g) x ( 1 mol/ 32g)= 0.71 mol O2 2. Determine mol fraction. (0.076 mol/ 0.79 mol)= 0.10 mol of halothane (0.71 mol/ 0.79 mol)= 0.90 mol of O2 3. Find Partial pressure: PA= XA + PT 0.10 x 862= 86 torr of Halothane 0.90 x 862= 776 torr of O2

Compare the ratio of effusion rates to the ratio of the diffusion rates for two gases with molecules of approximately the same mass and held under the same conditions. A.The ratio of the effusion rates is greater than the ratio of the diffusion rates. B.The ratio of the effusion rates is smaller than the ratio of the diffusion rates. C.The ratio of the effusion rates is equal to the ratio of the diffusion rates. D.The ratio of effusion rates is not related to the ratio of diffusion rates E.None of these

C. The ratio of the effusion rates is equal to the ratio of the diffusion rates.

Which of the following combinations lead(s) to a precipitation reaction? (i) Calcium nitrate(aq) + potassium chloride(aq) => (ii) Sodium chloride(aq) + lead(II) nitrate(aq) => (iii) Copper(II) chloride(aq) + sodium hydroxide(aq) => A.None lead to a precipitation reaction B.(i) and (iii) only C.(ii) only D.(ii) and (iii) only

D. (ii) and (iii) only Refer to and memorize solubility table and exceptions** Solve by: 1. Writing out the reactants and determining product, they will always be a double displacement reaction. After you have the product, balance the equation. Then look at each product and see if it is soluble or not. i) Ca(NO3)2 + 2 KCl -> CaCl2 + K2 (NO3)2 Halogens, nitrate, and group 1 metals are all soluble. ii) 2NaCl + Pb (II) (NO3)2 -> 2Na(NO3) + PbCl2 Na- group 1 and nitrate= soluble, but Lead is one of the exceptions that make halogens (Cl-) insoluble. iii) CuCl2 + Na(OH) -> Cu(OH)2 + NaCl OH- is insoluble, NaCl are soluble.

C2H4(g) + 3O2(g) => 2CO2(g) + 2H2O(l) ΔHf = -1411 kJ C(s) + O2(g) => CO2(g) ΔHf = -393.5 kJ H2(g) + 1/2O2(g) => H2O(l) ΔHf = -285.8 kJ Use the data given above to find the standard enthalpy of formation of ethylene, C2H4(g). A.731 kJ/mol B.87 kJ/mol C.1.41 x 103 kJ/mol D.52 kJ/mol E.2.77 x 103 kJ/mol

D. 52 kJ/mol 1411 + 2(-393.5) + 2(-285.5)= 52.4 kJ

A 24.1-g mixture of nitrogen and carbon dioxide is found to occupy a volume of 15.1 L when measured at 870.2 mmHg and 31.2C. What is the mole fraction of nitrogen in this mixture? A.0.539 B.0.500 C.0.461 D.0.426 E.0.574

E. 0.574 1. Find Total moles with the given data. n= ( 1.145x 15.1)/(0.082 x 304)= 0.69 mol total 2. I used a proportion to find the grams of N2 in 24.1 g. First I determined the molar mass of N2 (28g) and CO2 (44g), the sum being 72g. Our proportion is (28g/72g) = (x/24.1g) We'll solve for x= 9.37g N2 in our mixture. We'll divide this with the amount of mixture with our total amount. 9.37g/24.1g= 0.39 mol N2 0.39/0.69= 0.57 mole fraction of N2. *I'm not too entirely sure on the process but it worked.

An ideal gas (the system) is contained in a flexible balloon at a pressure of 1 atm and is initially at a temperature of 20.°C. The surrounding air is at the same pressure, but its temperature is 25°C. When the system has equilibrated with its surroundings, both systems and surroundings are at 25°C and 1 atm. In changing from the initial to the final state, which one of the following relationships regarding the system is correct? A.ΔE < 0 B.ΔE = 0 C.ΔH = 0 D.w > 0 E.q > 0

E. q > 0 Since the surroundings is giving heat to the system, that makes it endothermic, which is a +q that is >0.

Consider four identical 1.0-L flasks containing the following gases each at 25°C and 1 atm pressure. For which gas do the molecules have the greatest average kinetic energy? A.Ne B.Ar C.C2H6 D.Kr E.same for all the gases

E. same for all the gases *Refer to the 5 rules of kinetic molecular theory of gases The average kinetic energy is the same for all gases at a given temperature, regardless of the identity of the gas.

Automobile batteries use 3.0 M H2SO4 as an electrolyte. What volume of 1.20 M NaOH will be needed to neutralize 225 mL of battery acid?H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + Na2SO4(aq) A.0.045 L B.0.28 L C.0.56 L D.0.90 L E.1.1 L

E. 1.1 L Formula: (M1V1/n1)=(M2V2/n2) M1 and V1= Molarity and Vol. of acid n= moles of acid (use mole:mole ratio) M2V2= Molarity and vol. of base (3M)(.225 mL)/1 mol H2SO4= (1.20 M NaOH)(V2)/2 mol NaOH Solve for V2 (3 x .225 x 2)/1.20=1.1 L

Calculate q in J when 2.72 g of ice is cooled from -33.2.°C to -91.1.°C (c of ice = 2.087 J/g·K).

-329 ± 2%

How many sulfate ions are present in the following aqueous solution?458 mL of a solution containing 5.71 g/L aluminum sulfate. When you have the number, determine its log (base 10) and enter that value with 3 decimal places.

22.140 ± 0.05 Determine the correct formula so that its balanced. Al is +3 and SO4 has a -2 charge. For it to be balanced and neutral, we'll have Al2(SO4)3 that will dissolve completely to 2 Al + 3 SO4. Convert from: grams (given) of Al2(SO4)3 > mols of SO4 > atoms of SO4. (5.71g/1L) x ( 1 mol Al2(SO4)3/ 342g Al2(SO4)3) x (3mol SO4/ 1 mol Al2(SO4)3) x (.458 L/1)= 0.0229 mol *The .458 divided by is just a filler Now mols > atoms 0.0229 mol x (6.022 x 10^23)/1 mol= 1.37 x 10^22= log of the answer = 22.14

How many total moles of ions are released when the following sample dissolves completely in water? 1.51 mol K2SO4

4.53 ± 2% Group 1 metals are soluble so it will dissociate into the following: K2SO4 -> 2k + SO4 Where we can see our mole ratio of 1 K2SO4 mole: 2 mols K: 1 mol SO4. In total, K2SO4 dissociates into a total of 3 mols. Now we can convert. 1.51 mols K2SO4 x (3 mols total/ 1 mol K2SO4)= 4.53 mols of dissociated ions in our solution.

A sample of a gas in a piston/cyclinder apparatus occupies 912.3 mL at 3.22 atm and -23.34°C. If the volume of the sample is increased to 716.5 mL and the temperature changes to 11.42°C, what is the new pressure, in atm, of the sample? Do not enter units with your answer.

4.67 ± 2% PV=nRT, let's get the variable we want on one side for the initial sample of gas on the left side, and the after on the right side so it'll look like this: P1V1/T1=P2V2/T2 Rearrange it so we're solving for P2. (P1V1T2)/(T1V2)=P2 (3.22)(.9123)(284.42)/(249.66)(.7165)= 4.67

An unknown gas Q effuses 1.79 times as fast under the same conditions as Xe gas. What is the molar mass (g/mol) of Q? Enter your answer with one decimal place

41.0 ± 2% (R1/R2)= √(M2/M1) We don't know the rate of Xe, we just know the other compound is 1. 79 times as fast, so we'll just set the rate of Xe (R1) at 1. (1/1.79)=√(131/M1) (1^2 x 131)/(1.79^2)= 41.0 g/mol You can also rearrange the formula to be this, the outcome is the same but can be a little easier to manipulate. (R1^2 M1)= (M2 R2^2)

How many total moles of ions are released when the following sample dissolves completely in water? 1.86 x 10^25 formula units of strontium hydrogen carbonate. Enter your answer with one decimal place.

92.7 ± 2% Sr(HCO3)2= 3 ions when dissolved (1.86 x 10^25) x (1 mol/ 6.022 x 10^23) x (3 mols on ions/ 1 mol Sr(HCO3))= 92.7 mols of ions

How many grams of copper metal are produced from the reaction of 14.38 mL of 0.273 M Cu2+ with excess zinc metal? A.0.249 B.0.00393 C.0.499 D.3.35 E.0.00121

A. 0.249 Molarity= mole/L Here we have 0.273 moles/1 L of Cu2+ We gotta understand the equation incase there is more than 1 mole in the products side, here we have Cu+Zn= Cu+Zn. We want grams, so we need mols of Cu. We will get it from converting mols of Cu2+. (0.273 mol Cu2+/ 1 L) x ( 0.01438 L/ 1 mol) x (63 g Cu/ 1 mol)= 0.24 g Cu

A 140.0-g sample of water at 25.0°C is mixed with 120.9 g of a certain metal at 100.0°C. After thermal equilibrium is established, the (final) temperature of the mixture is 29.6°C. What is the specific heat capacity of the metal, assuming it is constant over the temperature range concerned? A.0.32 J/g°C B.0.63 J/g°C C.0.24 J/g°C D.3.2 J/g°C E.none of these

A. 0.32 J/g°C q=mcΔT, for 2 substances we'd have mcΔT=-mcΔT We'll put metal on the left side because it has a higher initial temperature. *I used 4.184 J/g for the specific heat (c) for water (120.9g)(c)(29.3-100)=-(140)(4.184)(29.6-25) (-8463)(c)=-2694 c=-2694/-8463=0.32 J/g

At the same temperature, the rate of effusion of neon atoms will be approximately _____________ times that of arsine gas molecules, AsH3. A.2.0 B.4.0 C.10.0 D.39.0 E.78.0

A. 2.0 Formula: (R1/R2)= √(M2/M1) Let's say the rate of effusion for AsH3 (R2) is 1, how many times greater will Ne be? Let's plug it into the formula. (R1/1)=√(78/20) Now solve for R1= 1.97=2.0

Calculate the number of moles of solute contained in 1.875 L of 1.356 M NaOH solution. A.2.543 mol B.1.383 mol C.0.7232 mol D.0.3932 mol E.0.001383 mol

A. 2.543 mol Molarity= moles/Liter moles= Moles x Liter

A mercury mirror forms inside a test tube by the thermal decomposition of mercury(II) oxide:2HgO(s) => 2Hg(l) + O2(g) ΔHrxn = 181.6 kJHow much heat is needed to decompose 555 g of the oxide? A.233 kJ B.-233 kJ C.465 kJ D.-465 kJ E.23.3 kJ

A. 233 kJ In our equation, 181.6 kJ is used to burn up 2 mols. But how many mols do we have in 555g? 1. Convert 555g x (1 mol HgO/ 216g HgO)= 2.6 mol HgO 2. Solve for kJ that will be used up, knowing that every 2 mol=181.6 kJ is burned. 2.6 mol x (181.6 kJ / 2 mol)= 236 kJ.

The pressure in a 12.2 L vessel that contains 2.34 g of carbon dioxide, 1.73 g of sulfur dioxide, and 3.33 g of argon, all at 42°C is _____ torr. A.263 B.134 C.395 D.116 E.0.347

A. 263 We're looking for overall pressure P=nRT/V. But what is n? We need to find the molecules of each first. 1. 2.34g/44g= 0.053 mol CO2 1.73g/ 64g= 0.027 mol SO2 3.33g/ 40g= 0.083 mol Ar Sum of all moles= 0.163 mols, now plug it in. Let's first, convert 0.082 (R, the constant) into torr. So it'll be 62.32 (0.163 x 62.32 x 315)/ 12.2= 262 *If you round the sum of moles to 0.16 like I did you'll get a smaller number but it's close to the answer/correct.

Aqueous copper(II) nitrate reacts with potassium iodide to yield solid copper(I) iodide, aqueous potassium nitrate, and solid iodine. The balanced net ionic equation for the reaction is: A.2Cu2+(aq) + 4I-(aq) --> 2CuI(s) + I2(s) B.K+(aq) + 3I-(aq) --> KI(s) + I2(s) C.Cu2+(aq) + I-(aq) --> CuI(s) + I2(s) D.K+(aq) + NO3-(aq) --> KNO3(s) E.None of these.

A. 2Cu2+(aq) + 4I-(aq) --> 2CuI(s) + I2(s) *Write out the correct formula. I got stuck on it for a long time before realizing solid iodine is 1 of the 7 diatomic elements. So it can only exist as I2. So our full balanced equation.... 2Cu(NO3)2 + 4KI -> 2CuI + 4KNO3 + I2 1. dissolve aqueous solutions into separate ions, leave mentioned solids as is. 2Cu + 4NO3 + 4K + 4I -> 2CuI + 4K + 4NO3 +2I 2. Cancel out ions that show up on both sides and you'll get answer A.

You mix 55 mL of 1.00 M silver nitrate with 25 mL of 1.06 M sodium chloride. What mass of silver chloride should you form? A.3.8 g B.7.6 g C.4.2 g D.8.4 E.none of these

A. 3.8 g We're looking for grams of AgCl. First, lets write out the formula so we can see the mole:mole ratio. We have AgNO3 + NaCl -> AgCl + NaNO3 at a 1:1 ratio of everything. Now we need to know how much AgCl we make, but we can only make a certain amount and that depends on the amount of Ag and Cl. So we need to find the limiting reactant that decides how much we can actually make. Lets convert everything to moles of Ag and moles of Cl, because that's what makes up AgCl and by finding the separate mole amounts we would be able to find the compounds mole amount. I'll write out everything just so understanding the stoichiometry of it all will be clear. (.055 L) x (1 mole AgCl/ 1 L) x (1 mol AgCl/ 1 mol AgNO3) x (1 mol Ag/1 mol AgCl)= .055 mol Ag (.025 L) x (1 mol NaCl/ 1 L) x (1 mol AgCl/ 1 mol NaCl) x (1 mol Cl/ 1 mol AgCl)= .027 mol Cl The most we can make is .027 moles of Cl, thus is our limiting reactant. So we can use that to find the amount of grams of AgCl. (.027 mol Cl) x (143.32 g AgCl/ 1 mol Cl)= 3.87g AgCl.

Calculate the heat capacity of a piece of iron metal if 3370 J of heat produces a temperature increase from 100°C to 200°C. A.33.7 J°C-1 B.3.37 x 105 J°C-1 C.0.0297 J°C-1 D.4.02 J °C-1 E.0.449 J°C-1

A. 33.7 J°C-1

What is the pressure in pascals of the gas in the flask below? *Picture shows a manometer with a closed end, where the right tube is higher, thus Pgas > Patm. The Δh=3.56 cm A.4.75 x 10^3 Pa B.4.57 x 10^2 Pa C.0.268 Pa D.3.59 Pa E.47.5 Pa

A. 4.75 x 10^3 Pa There's a difference in the closed and open ended manometer, is the open ended one has pressure from the atmosphere. That's when we add or subtract the height from the atmospheres pressure based on if the pressure of the gas or the atmosphere is greater. In a closed end manometer the Pgas is just the height, so we just need to convert it to Pa. Convert 3.56 cm to mm -> 35.6 mm 35.6mm x ( 1 atm/ 760 mm Hg) x (101325 Pa/ 1 atm)= 4.75 x 10^3

A mixture of three gases has a total pressure at 298 K of 1560 mm Hg. the mixture is analyzed and is found to contain 1.50 mol Ne, 2.65 mol Ar, and 1.75 mol Xe. What is the partial pressure of Xe? A.463 mmHg B.225 mmHg C.658 mmHg D.396 mmHg E.701 mm Hg

A. 463 mmHg Daltons Law Mole Fraction: x=mole fraction nA= moles of target element nT= total moles A=element (Xe) Formulas: PA= XA + PT So the partial pressure= mole fraction of the element + Total Pressure Mole fraction (XA) can be found by XA= (nA/nT) or (PA/PT) We can use either the moles/pressure of the target element divided by the total amount of moles/pressure. So in total we have 5.9 moles. Our mole fraction= 1.75 mol Xe/5.9 mols= 0.30 mol fraction of Xe. Use PA= XA + PT to find the partial pressure of Xe. 0.29 mols of Xe x 1560 mm Hg= 463 mm Hg

To prepare a 0.400 M NaCl solution, add _______ of NaCl and enough water to form a solution with a volume of _______ in a volumetric flask. A.5.84 g, 250 mL B.29.2 g, 500 mL C.58.4 g, 4.00 L D.11.7 g, 400 mL E.58.4 g, 1.00 L

A. 5.84 g, 250 mL

Which, if any, of the following is not a redox reaction? A.CO3^2- + HSO4- --> HCO3- + SO4^2- B.2Al(s) + 3H2SO4(aq) --> 3H2(g) + Al2(SO4)3(aq) C.2H2O --> 2H2 + O2 D.2NaI + Br2 --> 2NaBr + I2 E.All of them are redox reactions.

A. CO32- + HSO4- --> HCO3- + SO42-

NH4Br is very soluble in water. It is an example of a A.strong electrolyte B.non-electrolyte C.weak electrolyte D.weak acid E.precipitate

A. strong electrolyte

Which of the following solutions will be the poorest conductor of electrical current? A.sucrose, C12H22O11(aq) B.sodium chloride, NaCl(aq) C.potassium nitrate, KNO3(aq) D.lithium hydroxide, LiOH(aq) E.sulfuric acid, H2SO4(aq)

A. sucrose, C12H22O11(aq)

When 10 g samples of the following materials absorb 100 J of heat, the temperature increase indicated is observed. Which material has the largest specific heat? A.water, 2 K B.wood, 6 K C.copper, 26 K D.cement, 11 K E.glass, 12 K

A. water, 2 K

Estimate ΔE, in kJ, for the following reaction from the bond energies given below. N2(g) + 3H2(g) → 2NH3(g) Bond energy (kJ/mol): N-N, 163; N=N, 418; N≡N, 946; H-H, 436; N-H, 389. A.-81 B.+608 C.+863 D.+4588 E.+81

A. -81 Bonds broken-Bonds Formed 946 x 3(436) - 2 (3 x 389)= 2334 -2334= -80 kJ

Consider four identical 1.0-L flasks containing the following gases each at 25°C and 1 atm pressure. For which gas do the molecules have the highest average velocity? A.H2 B.O2 C.NH3 D.SO2 E.same for all gases

A. H2 Don't get it confused with average kinetic energy! We pick the lightest element, that will be the fastest.

SO2(g) + 1/2O2(g) => SO3(g) ΔH = -99.1 kJGiven the above data, calculate the enthalpy change ΔH when 89.6 g of SO2 is converted to SO3. A.-69.3 kJ B.-139 kJ C.69.3 kJ D.139 kJ E.-111 kJ

B. -139 kJ Convert from grams > mol using mole:mole ratio > kJ. (89.6 g) x (1 mol SO2/64g) x (-99.1 kJ/1 mol SO3)= -139

Calculate the molarity of F- in a 250.0 mL solution containing 18.34 g LiF and 4.32 g NaF. A.0.810 B.3.24 C.0.202 D.4.77 E.0.0906

B. 3.24 Molarity= mols/L, we need moles so convert the grams > moles 18.34g LiF x (1 mol/26 g LiF)= 0.71 mol LiF 4.32g NaF x (1 mol/42g NaF)= 0.10 mol NaF (.71 + .10)/.25 L= 3.24 M

What is the heat capacity of a metal slug if a temperature rise from 25C to 54C requires 135 J? A.0.21 J/C B.4.7 J/C C.0.60 J/C D.2.5 J/C E.-4.6 J/C

B. 4.7 J/C q=mcΔT; but they don't give us mass so it'll be q=cΔT where we solve for c=q/ΔT 54 C - 25 C= 29 C c= (135/29)= 4.7 J/C

A student wants to make 100 mL of 0.30 M HCl. How many milliliters of 6.0 M HCl is needed to make this solution? A.200 B.5 C.500 D.50 E.10

B. 5 Formula for dilution: m1v1=m2v2 -> (0.3)(100)=(6)(v2) > (0.3 x 100)/6= 5 mL

One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.41 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T = 300 K; 1 L·atm = 101.3 J) A.375 J B.751 J C.2.25 × 103 J D.852 J E.none of these

B. 751 J Work done under constant external pressure W = P∆V W = P(Vf—Vi) W = 1 atm × (8.41—1) litre W = 1 atm × 7.41 litre= 7.41 Litre•atm Given that, 1 Litre•atm = 101.3J W=(7.41 Litre•atm) x (101.3J/1 Litre•atm) W ≈ 751J

Which of the following indicates the greatest pressure? A.1 atm B.777 torr C.5.5 mm Hg D.1000 Pa E.12 psi

B. 777 torr Convert all to atm 1 atm= 760 torr= 760 mm Hg=101.3 kPa= 14.7 psi

Two iron bolts of equal mass-one at 100. °C, the other at 56°C-are placed in an insulated container. Assuming the heat capacity of the container is negligible, what is the final temperature inside the container (c of iron = 0.450 J/g·K) A.155°C B.78°C C.44°C D.88°C E.None of these is within 5% of the correct answer

B. 78°C mCΔT, but with no mass. Plug in the values and solve for Tf. -(0.450)(Tf-100)= (0.450)(Tf-56) -0.450Tf + 45 = 0.450Tf + 25.2 -.450Tf - 0.450Tf= -25.2 - 45 -0.90Tf = -70 Tf= 78 C

How many liters of CO2 at 0 °C and 380 torr are obtained in the reaction of 3 g of ethane (C2H6) with excess O2? A.13 L B.9 L C.4.5 L D.3 L

B. 9 L You would need to the compound ethane (I added it in), and the overall balanced formula. Here's the balanced formula: 2C2H6 + 7O2 -> 4CO2 + 6H2O Find number of moles of CO2, by converting grams of ethane to mols > using a mole to mole ratio to find mols of CO2. 3g C2H6 x (1 mol/30g) = 0.10 mol ethane 0.10 mol x (4mol CO2/ 2 mol) = 0.20 mol CO2 Plug in all into V= nRT/P (0.20)(0.082)(273) / 0.5 = 8.95 L = 9L

When an evacuated 63.8 mL glass bulb is filled with an unknown gas at 22°C and 747 mmHg, the bulb gains 0.103 g in mass. Which of these could be the gas? A.Ne B.Ar C.N2 D.O2 E.C2H6

B. Ar Convert to atm. (747 mm Hg) x (1 atm/ 760 mm Hg)= 0.98 atm Plug in n=PV/RT n=(0.98 x 0.0638) / (0.082 x 295) = 0.0026 mol of unknown gas Remember molar mass is grams/mol, which we have both of! MM= (0.103 g/ 0.0026 mol)= 39.6g= 40g= Ar

For the formation of 1 mol of sodium chloride from its elements (heat is released) ΔH is positive. A.True B.False

B. False

For the freezing of liquid water Δ H is positive. A.True B.False

B. False

A spectator ion accepts protons in a reaction. A.True B.False

B. False A spectator ion is an ion that does not take part in the chemical reaction and is found in solution both before and after the reaction

A closed end manometer is usually used to measure atmospheric pressure. A.True B.False

B. False Manometer measures pressue of gas in a container.

What is the effect of the following on the volume of 1 mol of an ideal gas?The pressure changes from 760 torr to 202 kPa and the temperature changes from 37.0°C to 155 K. A.Volume of gas does not change B.New volume is 25% of the original volume C.New volume is 400% of the original volume D.New volume is 200% of the original volume E.New volume is 50% of the original volume

B. New volume is 25% of the original volume

Which of the following is the correct net ionic equation for the reaction that occurs when aqueous solutions of Pb(ClO3)2 and Na2SO4 are mixed? A.ClO3-(aq) + Na+(aq) --> NaClO3(s) B.Pb2+(aq) + SO42-(aq) --> PbSO4(s) C.Pb(ClO3)2(aq) + Na2SO4(aq) --> PbSO4(s) + NaClO3(aq) D.Pb2+(aq) + 2ClO3-(aq) + 2Na+(aq) + SO42-(aq) --> PbSO4(s) + 2Na+(aq) + SO42-(aq) E.none of the above

B. Pb2+(aq) + SO42-(aq) --> PbSO4(s) *Memorize the exceptions to soluble and insoluble compounds. SO4 is usually soluble EXCEPT when with Pb (lead). So that makes it a solid and we won't dissolve it in our formula. So after determining the right product and balancing it, we'll get... Pb + (ClO3)2 + Na2 + SO4 -> PbSO4 + 2Na + 2 ClO3 Cancel out ones that show up on both sides... Pb + SO4 -> PbSO4

Carbon dioxide deviates more from the ideal gas law than diatomic hydrogen partly because: A.carbon dioxide is more reactive B.carbon dioxide has a larger molar mass and so has greater intermolecular forces C.hydrogen has greater intermolecular forces D.all of the above

B. carbon dioxide has a larger molar mass and so has greater intermolecular forces

Select the classification for the following reaction: H2CO3(aq) → H2O(l) + CO2(g) A.combination B.decomposition C.displacement D.acid-base E.None of these choices is correct.

B. decomposition

Select the classification for the following reaction.2I¯(aq) + Cl2(aq) → I2(aq) + 2Cl¯(aq) A.combination B.displacement C.decomposition D.precipitation E.acid-base

B. displacement

Given a 1.00 g sample of each of the following metals at 50°C, which would raise 100 g of water at 25°C to the highest temperature? A.iron (sp heat = 0.449 Jg-1°C-1) B.stainless steel (sp heat = 0.50 J g-1°C-1) C.lead (sp heat = 0.128 J g-1°C-1) D.copper (sp heat = 0.385 J g-1°C-1) E.silver (sp heat = 0.235 J g-1°C-1)

B. stainless steel (sp heat = 0.50 J g-1°C-1)

What volume, in L, of 10.0 M HCl is needed to make 2.00 L of 2.00 M HCl solution by dilution with water? Assume volumes are additive. A.0.800 L B.0.400 L C.0.200 L D.0.100 L E.None of these choices is correct.

B. 0.400 L Formula: M1V1=M2V2 M1= 10 V1= ? M2= 2 V2= 2 Plug in and solve for V1 V1= (2 x 2)/10= 0.400 L

Calculate the molarity of 1.25 L of solution which contains 238.7 g of cobalt(II) chloride. A.2.30 M B.1.47 M C.0.435 M D.0.680 M E.1.25 M

B. 1.47 M Molarity=moles/L Determine compound> CoCl2 *The roman numerals for metals is basically their charge. Convert grams>moles (238.7g x (1 mol/129.83g)= 1.84 mol M=(1.84mol/1.25L)=1.47 M of CoCl2

A gas mixture, with a total pressure of 300. torr, consists of equal masses of Ne (atomic weight 20.)and Ar (atomic weight 40.). What is the partial pressure of Ar, in torr? A.75 torr B.100. torr C.150. torr D.200. torr E.None of these choices is correct.

B. 100. torr

What volume of concentrated nitric acid (15.0 M) is required to make 300 mL of a 2.5 M nitric acid solution? A.1.8 L B.50 mL C.12.5 mL D.18 mL E.8 mL

B. 50 mL M1V1=M2V2 M1= 15M V1=?? M2= 2.5M V2= 300 mL Plug in and solve for V1. 15 x V1= 300 x 2.5 V1= (300 x 2.5)/15 V1=50mL

A 0.00100 mol sample of Ca(OH)2 requires 25.00 mL of aqueous HCl for neutralization according to the reaction below. What is the concentration of the HCl? Equation: Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) A.0.0200 M B.0.0400 M C.0.0800 M D.4.00 × 10-5 M E.None of these choices is correct.

C. 0.0800 M concentration=Molarity=moles/Liters, we'll take the mols of HCl and use the mole ratio in the equation. Mole ratio of: 1 mole of Ca(OH)2 to 2 moles of HCl. Need mols of HCl, we have 0.001 mol Ca(OH)2 x (2 mol HCl/1 mol)= 0.002 mol HCl Convert mL > L 25 mL x (1 L/1000 mL)= .025 L Molarity= 0.002 mol HCl/.025 L= 0.08M

If 0.750 L of argon at 1.50 atm and 177°C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0°C are added to a 1.00-L flask and the temperature is adjusted to 25.0°C, what is the resulting pressure in the flask? A.0.0851 atm B.0.244 atm C.0.946 atm D.1.74 atm E.1.86 atm

C. 0.946 atm We're looking for total pressure, we have everything except the total amount of moles. First find the moles of Ar and SO2, that is our n in PV=nRT formula. *Before you start convert everything to atm or K. 1. n=PV/RT nAr= (1.50 x .750)/(0.082 x 450)= 0.030 mol Ar. nSO2= (.94 x 0.235)/(0.082 x 336)= 0.008 mol SO2. So total moles= 0.030 + 0.008= 0.038 mol 2. Plug into P=nRT/V -> (0.038 x 0.082 x 298)/1= 0.93 atm

Which of the following will be least soluble in water? A.potassium sulfate, K2SO4 B.ammonium nitrate, NH4NO3 C.chloromethane, CH3Cl D.calcium chloride, CaCl2 E.ethanol, C2H6O

C. chloromethane, CH3Cl LEAST soluble, so it's still soluble. A. Both are soluble= strong B.Both are soluble= strong C. CH3= not insouble, but not exactly soluble=weak, Cl-= soluble D.CaCl2= both soluble= strong E. C2H6O= OH- is insoluble, so it does not count!

Gases generally have A.no decrease in volume when pressure is increased B.no increase in volume when temperature is increased C.low density D.closely packed particles E.high density

C. low density A. Increase in pressure= decrease in volume (Boyle's Law) B. PV=nRT if you get stuck! V=T so they increase together! C. They're not dense like a solid or liquid. D. Particles are moving and don't have defined shape like solids E. Refers to Solids.

In a mixture of helium and chlorine, occupying a volume of 14.6 L at 871.7 mmHg and 28.6C, it is found that the partial pressure of chlorine is 355 mmHg. What is the total mass of the sample? A.50.6 g B.1.10 g C.21.1 g D.1.60 g E.19.5 g

C. 21.1 g Lets convert everything so it's easier to work with. Ptotal= 871.7=1.15 atm Pcl= 355=0.47 atm T= 28.6C=301 K V= 14.6L 1. Work Backwards. What are we looking for? We need the # of moles of He and Cl to find the total mass. We can use our Partial Pressure formula> PA= XA + PT , but instead of total pressure we can use total # of moles since they can be interchangable when temp and vol are the same. But we first need the total amount of moles and mole fraction of Cl and He. 2. n= (PV/RT) > (1.15 x 14.6)/(0.082 x 301)= 0.68 mol total Now we can solve for our mole fraction, it's the rratio of moles of our element to the mixtures moles. 3. (0.47/1.15)= 0.41 Cl; Remember that the partials of pressure/mols all equal to 1, so we can do 1- 0.41= 0.59 He. 4. They have no units because it's a ratio, if we use PA= XA + PT with the total amount of moles instead of pressure, we will end up with the partial amount of moles in the mixture. (0.41 x 0.68 mol)= 0.28 mol Cl2 (0.59 x 0.68 mol)= 0.40 mol He 5. Convert to grams and add together. 0.28 mol Cl x (71g/1 mol)= 19.9g 0.40 mol He x (4g/1mol)= 1.6 g 1.6 + 19.9= 21.5g

What is the coefficient of H2O after balancing the following equation? Bi^3+(aq) + Fe^3+(aq) + H2O => BiO3^1- + Fe2^+ + H^+(aq) A.1 B.2 C.3 D.4 E.None of these

C. 3 1. Identify which is compound/element is getting reduced and oxidized. Bi goes from +3 > +5 and is being oxidized and Fe goes from +3 > +2 and is being reduced. 2. We'll separate them in to half reactions Bi^+3 + H2O-> BiO3 and Fe^+3 -> Fe^+2 3. Balance out Oxygens and hydrogens with H2O and H+, respectively. Bi^+3 + 3H2O -> BiO3 + 6H+ the other equation is already balanced, but we will return to it later. 4. Add e- to balance out the charges in the reactants and products in each half reaction. Bi^+3 + 3H2O -> BiO3 + 6H+ + 3 e- We added 3 electrons, because the reactants have an overall charge of +3 from Bi, (H2O is neutral). On the products side we had +6 (from H+) so we'll add +3 e- to lower it to +3. Now they're balanced. 5. Multiply the other half reaction by the # of electrons to make the # of electrons equal in each half reaction. We'll just multiply the whole half reaction by 3. 3(Fe^+3 -> Fe^+2) is now 3Fe^+3 -> 3Fe^+2 6. Add it all together We probably didn't need to go this far, since we already know the # of H2O molecules. But it's good practice. Bi^+3 + 3H2O + 3Fe^+3 -> BiO3 + 3Fe^+2 + 6H+

A mixture of three gases has a pressure at 298 K of 1380 mm Hg. The mixture is analysed and is found to contain 1.27 mol CO2, 3.04 mol CO, and 1.50 mol Ar. What is the partial pressure of Ar? A.238 mm Hg B.302 mm Hg C.359 mm Hg D.1753 mm Hg E.8018 mm Hg

C. 359 mm Hg partial pressure of Ar= mole fraction x total fraction PAr= nAr x PT mole fraction of Ar= (1.50 x 5.81)= 0.26 Partial Pressure= 0.26 x 1380= 359 mm Hg

A bomb calorimeter has a heat capacity of 783 J/C and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How much heat, in kilojoules, is evolved or absorbed by a reaction when the temperature goes from 23.73 C to 26.01 oC? c of H2O = 4.184 J/g.K A.4200 kJ absorbed B.2420 kJ absorbed C.4.21 kJ evolved D.2.42 kJ absorbed E.1.78 kJ evolved

C. 4.21 kJ evolved mcΔT=mcΔT; But the calorimeter doesn't need the mass so it's cΔT=mcΔT. where we'll have q= heat absorbed by calorimeter + heat absorbed by water so our full formula is: q=cΔT + mcΔT ΔT= 26.01 - 23.73= 2.28 C Plug it all in: (783)(2.28) + (254)(4.184)(2.28)= 4208 J Convert to kJ: 4208 J x (1 kJ/ 1000J)= 4.21 kJ

If 25.0 g of water at 21C is mixed with 45.0 g of water at 75C, what is the final temperature of the mixture? A.40C B.48C C.56C D.63C E.38C

C. 56C q=mcΔT where mcΔT=-mcΔT but c is the same here because it's both water so we are left with mΔT=-mΔT m1(Tf-Ti)= -m(Tf-Ti) 25(Tf-21)=-45 (Tf-75) FOIL 25Tf-525=-45Tf + 3375 25Tf + 45Tf= 3375 + 525 70Tf= 3900 Tf= 56 C

Complete combustion of 1.0 metric ton of coal (assuming pure carbon) to gaseous carbon dioxide releases 3.3 x 10^10 J of heat. Convert this energy to kilocalories A.7.9 x 10^9 kcal B.1.4 x 10^8 kcal C.7.9 x 10^6 kcal D.1.4 x 10^11 kcal E.None of these is within 5% of the correct answer

C. 7.9 x 106 kcal Convert Joules > kcal *4.184 J=1 cal *1000 cal=1 kcal *1 kJ=1000J (3.3 x 10^10 J) x (1 cal/ 4.184 J) x (1 kcal/ 1000cal)= 7.9 x 10^6 kcal

A flask with a volume of 3.16 L contains 9.33 grams of an unknown gas at 32.0°C and 1.00 atm. What is the molar mass of the gas? A.7.76 g/mol B.66.1 g/mol C.74.0 g/mol D.81.4 g/mol E.144 g/mol

C. 74.0 g/mol Where moles=n=(mass of sample/Molar Mass) OR n=m/MM So we keeping that in mind and knowing n=PV/RT, we can say that m/MM=PV/RT or MM=mRT/PV. (9.33g)(0.082)(305)/(1)(3.16)= 73.84 g/mol *Keep track of units. Remember the unit of our contstant (R)= 0.082 atm x L/mol x K.

Which one of the following is not a redox reaction? A.2H2O2(aq) → 2H2O(l) + O2(g) B.N2(g) + 3H2(g) → 2NH3(g) C.BaCl2(aq) + K2CrO4(aq) → BaCrO4(aq) + 2KCl(aq) D.2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) E.2H2O(g) → 2H2(g) + O2(g)

C. BaCl2(aq) + K2CrO4(aq) → BaCrO4(aq) + 2KCl(aq)

Select the net ionic equation for the reaction between lithium hydroxide and hydrobromic acid. LiOH(aq) + HBr(aq) → H2O(l) + LiBr(aq) A.LiOH(aq) → Li+(aq) + OH¯(aq) B.HBr(aq) → H+(aq) + Br¯(aq) C.H+(aq) + OH¯(aq) → H2O(l) D.Li+(aq) + Br¯(aq) → LiBr(aq) E.Li+(aq) + OH¯(aq) + H+(aq) + Br¯(aq) → H2O(l) + LiBr(aq)

C. H+(aq) + OH¯(aq) → H2O(l) 1. Dissolve/separate all ions that can dissolve. Li + OH + H + Br -> H2O + Li + Br 2. Cancel out the elements that are the same on both sides of the reaction. The remaining equation is the net ionic equation. OH + H -> H2O

Select the net ionic equation for the reaction between lithium hydroxide and hydrobromic acid. LiOH(aq) + HBr(aq) → H2O(l) + LiBr(aq) A. LiOH(aq) → Li+(aq) + OH¯(aq) B. HBr(aq) → H+(aq) + Br¯(aq) C. H+(aq) + OH¯(aq) → H2O(l) D. Li+(aq) + Br¯(aq) → LiBr(aq) E. Li+(aq) + OH¯(aq) + H+(aq) + Br¯(aq) → H2O(l) + LiBr(aq)

C. H+(aq) + OH¯(aq) → H2O(l) How to Solve: Dissolve everything on both sides of the reaction that can dissolve- so all aqueous compounds. Li + OH + H + Br -> H2O + Li + Br Cancel out everything that shows up on either side. OH + H -> H2O is the net ionic equation.

Which of the following do you need to know to be able to calculate the molarity of a salt solution? I. the mass of salt added II. the molar mass of the salt III . the volume of water added IV. the total volume of the solution A.II, III B.I, II, III C.I, II, IV D.You need all of the information. E.I, III

C. I, II, IV

Calculate the molarity of sulfate ion in a solution that is both 0.186 M Na2SO4 and 0.0560 M Al2(SO4)3 A.0.307 B.0.484 C.0.354 D.0.242 E.0.614

C. 0.354 Remember that Molarity=moles/L So we can say that we have 0.186 mol/1L of Na2SO4 and 0.0560 mol/1L of Al2(SO4)3. Now we can convert them to moles of SO4 using a mole:mole ratio. (0.186 moles/1L) x ( 1 mol SO4/1 mol Na2SO4)= 0.186 M SO4 (0.0560 moles/1L) x (3 mol SO4/1 mol Al2(SO4)3)= 0.168 M SO4 Add the 2 together.

Balance the following equation :__HCl(aq) + __FeCl2(aq) + __H2O2(aq) => __FeCl3(aq) + __H2O(l) A.2HCl(aq) + FeCl2(aq) + H2O2(aq) =>FeCl3(aq) + 2H2O(l) B.HCl(aq) + FeCl2(aq) + H2O2(aq) =>FeCl3(aq) + 2H2O(l) C.2HCl(aq) + 2FeCl2(aq) + H2O2(aq) =>2FeCl3(aq) + 2H2O(l) D.2HCl(aq) + 2FeCl2(aq) + H2O2(aq) =>2FeCl3(aq) + H2O(l) E.None of these

C. 2HCl(aq) + 2FeCl2(aq) + H2O2(aq) =>2FeCl3(aq) + 2H2O(l)

Which of the following substances is not soluble in water? A.HCl B.Ca(NO3)2 C.PbSO4 D.(NH4)2SO4 E.KOH

C. PbSO4

Among the first atoms in each of the following formulas, which one has the highest (most positive) oxidation number? A.CO3^2- B.NO3^- C.SO3 D.ClO^3- E.P2O5

C. SO3 **Only looking at the first atom. But I'll list out all of them. a) CO3 -2 > +4 on C and -6 for O for a total -2 charge. b)NO3- > +5 on N and -2 on O for a total -1 charge. c) SO3 > +6 on S and -6 on O for a neutral 0 charge. d) ClO -3 > +5 on Cl and -2 on O for a total -3 charge. e) P2O5 > +10 on P and -10 on O for a total neutral charge. We don't pick e because we're just looking at the 1st P atom, which just as +5 by itself.

A chunk of aluminum at 91.4°C was added to 200.0 g of water at 15.5°C. The specific heat of aluminum is 0.897 J/g°C, and the specific heat of water is 4.18 J/g°C. When the temperature stabilized, the temperature of the mixture was 18.9°C. Assuming no heat was lost to the surroundings, what was the mass of aluminum added? A.243 g B.34.7 g C.41.7 g D.43.7 g E.none of these

D. 43.7 g -mCΔT=mcΔT -m(0.897)(18.9-91.4)= (200)(4.18)(18.9 - 15.5) m65.03 = 2842.4 m= 43.71 g

The chemistry of nitrogen oxides is complex. Given the following reactions and their standard enthalpy changes, (1) NO(g) + NO2(g) => N2O3(g) ΔH = -39.8 kJ (2) NO(g) + NO2(g) + O2(g) => N2O5(g) ΔH = -112.5 kJ (3) 2NO2(g) => N2O4(g) ΔH = -57.2 kJ (4) 2NO(g) + O2(g) => 2NO2(g) ΔH = -114.2 kJ (5) N2O5(g) => N2O5(s) ΔH = -54.1 kJ calculate the heat of reaction for N2O3(g) + N2O5(s) => 2N2O4(g) A.-377.8 kJ B.35.0 kJ C.435.0 kJ D.-22.2 kJ E.None of these is within 5% of the correct answer.

D. -22.2 kJ 1) flip it to be ΔH = +39.8 kJ 5) flip it to be ΔH = +54.1 kJ 2) Flip it be ΔH = +112.5 kJ 3) Multiply by 2 because it's in our final equation ΔH =2*(-57.2 kJ) 4) ΔH = -114.2 kJ Sum of ΔH's = (39.9 + 54.1 +112.5 + 2*(-57.2) -114.2) kJ = -22.1

A sample of N2(g) was collected over water at 25C and 730 torr in a container with a volume of 340 mL. The vapor pressure of water at 25C is 23.76 torr. What mass of N2 was collected? A.0.58 g B.0.72 g C.0.91 g D.0.36 g E.0.080 g

D. 0.36 g We need pressure of N2, so to find it we'll subtract the vapor pressure of water from the total pressure to find the partial pressure of N2. 730-23.76= 706.24 toff N2= 0.929 atm. Convert everything to L and K. 340 mL= .340 L 25C=298 K We're looking for grams, which we can convert to with moles by using PV=nRT > n= PV/RT n=(0.929 x .34)/(0.082 x 298)= 0.0129 mols of N2 Now we can convert to grams. 0.0129 mols x (28g / 1mol)= .36g N2

The partial pressures of CH4, N2, and O2 in a sample of gas were found to be 151 mmHg, 511 mmHg, and 587 mmHg, respectively. Calculate the mole fraction of oxygen. A.21.0 B.0.359 C.0.409 D.0.470 E.0.772

D. 0.470 Mole fraction (XA) can be found by XA= (nA/nT) or (PA/PT) We can use either the moles/pressure of the target element divided by the total amount of moles/pressure. A=compound we're looking for T= Total n=moles X=Mole Fraction XO2= PO2/PT -> (587/1249)= 0.47

Consider a 1.00-L solution containing 85.5 g Al2(SO4)3 (FW = 342.15) and 21.3 g Na2SO4 (FW = 142.06). What are the molar concentrations of aluminum, sodium, and sulfate ions, respectively? A.0.25, 0.15 and 0.40 M B.0.50, 0.75 and 0.15 M C.0.50, 0.75 and 0.90 M D.0.50, 0.30, 0.90 M E.0.25, 0.30, 0.15

D. 0.50, 0.30, 0.90 M Find the # of moles in each compound using a mole:mole ratio. We don't have to plug it back into the molarity formula because the Volume is just 1. Al2(SO4)3 is 1 molecule with 2 molecules of Al and 3 molecules of SO4. Na2SO4 is 1 molecule with 2 molecules of Na and 1 molecule of SO4. 85.5g Al2(SO4)3 x (1 mol Al2(SO4)3 /342.15g) x (1 mol Al/1 mol)= .50 mol Al 85.5g Al2(SO4)3 x (1 mol Al2(SO4)3 /342.15g) x (3 mol SO4/1 mol)= .75 mol SO4 21.3g Na2SO4 x (1 molNa2SO4/ 142.06g) x (2 mol Na/1 mol)= .30 Na 21.3g Na2SO4 x (1 molNa2SO4/ 142.06g) x (1 mol SO4/1mol)= .15 SO4 .75 + .15 SO4= .90 mol SO4

What volume of a 0.100 M solution of HCl is required to neutralize a solution prepared by dissolving 10.0 g of KOH in 250 ml of H2O? A.0.178 L B.0.25 L C.0.71 L D.1.78 L E.178 L

D. 1.78 L M1V2=M2V2 Convert grams to moles then use it and .250 L to find the Molarity of KOH. 10g KOH x ( 1mol KOH/56.09g KOH) = .178 mol KOH Molarity= moles/Volume (.178 mol KOH/ .250 L )= .71 M KOH Plug it all in and solve for V1 (0.100 M)(V1)=(.71 M)(.25L) M cancels out and we're left with Liters. (.71M)(.25L)/(0.100 M)= 1.78 L

How many mL of concentrated nitric acid (HNO3, 16.0 M) should be diluted with water in order to make 2.00 L of 2.00 M solution? Assume volumes are additive. A.32.0 mL B.62.5 mL C.125 mL D.250. mL E.500. mL

D. 250. mL Equation: M1V1=M2V2 M1= Mass= 16 V1=Volume= ?? M2= 2 V2= 2 Plug in and solve for V1: 16 x V1 =2 x 2 ---> V1= (2x2)/16 > .25L > 250 mL

Data: 2Ba(s) + O2(g) => 2BaO(s) ΔH = -1107.0 kJ Ba(s) + CO2(g) + 1/2O2(g) => BaCO3(s) ΔH = -822.5 kJ Given the data above, calculate ΔH for the reaction below. Reaction: BaCO3(s) => BaO(s) + CO2(g) A.-1929.5 kJ B.-1376.0 kJ C.-284.5 kJ D.269.0 kJ E.537 kJ

D. 269.0 kJ Flip 2nd rxn to make BaCO3 to be on on reactant side: BaCO₃(s) ⇒ Ba(s) + CO₂(g) + 1/2O₂(g) ΔH = 822.5 kJ The first reaction has too many moles so we can divide it by 2, we can do this by just multiplying the ΔH by .5. Ba(s) + 1/2O₂(g) ⇒ BaO(s) ΔH = -553.5 KJ Add both reactions and cancel an reactants and products are the same on either side. BaCO₃(s) ⇒ BaO (s) + CO₂ (g) (822.5 - 553.5)= 269=ΔH

What are the coefficients in front of C2H6O and C2H4O when the equation given below is balanced? The reaction occurs in aqueous acid.Equation: C2H6O + Cr2O7^2- + H^+ => C2H4O + H2O + Cr^3+ A.1, 1 B.2, 2 C.2, 3 D.3, 3 E.6, 6

D. 3, 3 Determine oxidation numbers and which one is being oxidized and reduced. C has a oxi. # of -2 in C2H6O and loses e- and goes to -1 in C2H4O Cr has a oxi. # of +6 in Cr2O7 and gains e- and goes to +3 in Cr^+3. 1. Separate them into half reactions, include H+ and H2O so you won't get forget them when you add them later on during balancing. C2H6O + H+ -> C2H4O + H2O Cr2O7^-2 -> Cr^+3 2. Balance out elements that aren't H or O. Then Balance out O by adding H2O and balance out H by adding H+. C2H6O + 2H+ -> C2H4O + 2H2O 2H+ + Cr2O7^-2 -> 2Cr^+3 + H2O 3. Determine the charge of products and reactants for each half-reaction. Then add e- to either side so they are balanced. 2e- + C2H6O + 2H+ -> C2H4O + 2H2O; 2e- cancels out the 2H+ so it's overall neutral. 2H+ + Cr2O7^-2 -> 2Cr^+3 + H2O + 3e-; reactant side is neutral, we'll add 3 e- to balance out the Cr^+3 for an overall neutral charge. 4.Multiply each half reaction so the e- are equal. So we can multiply the 1st rxn. by 3 and the 2nd rxn. by 2 so they'll have 6e-. 6e- + 3C2H6O + 6H+ -> 3C2H4O + 6H2O 4H+ + 2Cr2O7^-2 -> 4Cr^+3 + 2H2O + 6e- Add them all together and cancel out anything that shows up on either sides. We already got our answer, I just took all the way to the end for future reference.

A 200. g sample of a metal is heated to 54 degrees Celsius and then dropped into 119. g of water in a calorimeter. The temperature of the water rises from 21.0 degrees Celcius to 41.0 degrees Celcius. What is the specific heat of the metal (J/g.K)? A.91.5 B.21.9 C.0.108 D.3.83 E.-3.83

D. 3.83 -mcΔT=mcΔT I'll put metal on the left side because it's at a higher temperature and losing heat. I'll move the negative to whatever is losing heat. ***To remember= specific heat of water= 4.184 -(200)(c)(41-54) = (119)(4.184)(41-21) Solve for c 2600c= 9958 c= 3.83

A 0.850-mole sample of nitrous oxide, a gas used as an anesthetic by dentists, has a volume of 20.46 L at 123°C and 1.35 atm. What would be its volume at 468°C and 1.35 atm? A.5.38 L B.10.9 L C.19.0 L D.38.3 L E.77.9 L

D. 38.3 L PV=nRT; charles law= V1/T1=V2/T2 Where pressure is the same and moles are the same so that leaves us with volume and temperature. Knowing the ideal gas law makes it so there's less to memorize. Solving for V2. (V1 x T2)/T1= (20.46)x(741)/(396)= 38.3 L

The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium azide, NaN3, which decomposes when activated by an electronic igniter to produce nitrogen gas that fills the bag. How many liters of nitrogen, measured at 25 °C and 1.00 atm, will be produced by 100.0 g of NaN3?Equation: 2NaN3(s) --> 2Na(s) + 3N2(s) A.1.33 L B.15.9 L C.37.6 L D.56.4 L E.205 L

D. 56.4 L V=nRT/P We don't have mols, so we'll find mols of N2 in respect to the mole:mole ratio of NaN3:N2. molar mass of NaN3= 65g 100 g x (1 mol NaN3/65g NaN3)= 1.54 mol NaN3 Find mols of N2 using mole:mole ratio. 1.54 mol NaN3 x (3 mol N2/ 2 mol NaN3)= 2.31 mol N2 Plug it in and solve. (2.31)(0.082)(298)/1.00)= 56.4 L

If excess zinc metal is added to 100 mL of a solution containing 0.70 M AgNO3, how many grams of metal will be displaced? A.3.8 B.9.1 C.0.070 D.7.6 E.5.6

D. 7.6 We only count for the grams of Ag, because it will be displaced by adding Zinc. We know 0.70M AgNO3= 0.70 mols/1 Litre. We already know the mole:mole ratio> 1:1. .1 L x (0.7 mol/ 1 L)= 0.07 mol Convert to grams 0.07 mol AgNO3 x (107.9 g/ 1 mol)= 7.6g Ag

What is the ratio of the average rate of effusion of NOF(g) to that of HBr(g) at 400 K? A.81:49 B.7:9 C.49:81 D.9:7 E.The average rate of effusion is the same for the two gases.

D. 9:7 Average rate of effusion. Know how the rate of effusion formula is set up where the masses are opposite of the rates. (R1/R2)= √(M2/M1) We don't know the rates, but we can determine the molar masses (M) and that will be the ratio of our average rate. NOF= 49 HBr= 81 (R1/R2)= √(81/49)= 9/7

Consider 0.1 M solutions of the following substances. Which would have the greatest electrical conductivity? A.NH3 B.CH3COOH C.HCl D.Ba(OH)2 E.NH4Br

D. Ba(OH)2

Which one of the following substances would you expect to be a nonelectrolyte in aqueous solution? A.HBr B.CH3CH2COOH C.KI D.CH3CH2OH E.Ca(NO3)2

D. CH3CH2OH

A sample of a gaseous compound of nitrogen and oxygen weighs 5.25 g and occupies a volume of 1.00 L at a pressure of 1.26 atm and temperature of -4.0 °C. Which of the following molecular formulas could be that of the compound? A.NO B.NO2 C.N3O6 D.N2O4 E.N2O5

D. N2O4

What are the major species in solution when solid ammonium nitrite is dissolved in water? A.NH3, NO3- B.NH4+, HNO3- C.NH3, NO2- D.NH4+, NO2- E.NH4+, NO3-

D. NH4+, NO2- Refer to list of common polyatomic ions and solubility rules/chart

When its applied to a real gas, the ideal gas law tends to become inaccurate when A.the temperature is raised above the temperature of STP. B.the pressure is lowered and molecular interactions become significant. C.large gas samples (large n) are involved. D.the pressure is raised and the temperature is lowered. E.the volume expands beyond the standard molar volume.

D. the pressure is raised and the temperature is lowered. Ideal gases behave best at low pressures and high temperature

What is the oxidation number of S in Na2S2O6? (assume both sulfurs have the same ox. no.) A.+4 B.+3 C.+6 D.+5 E.+2

D. +5

Neutralization of 18.02 mL H2SO4(aq) required 13.14 mL of 0.35 M NaOH(aq). What is the molar concentration of H2SO4(aq)? A.0.26 B.0.0030 C.0.96 D.0.13 E.0.48

D. 0.13 1.Write out and balance the reaction. H2SO4 + 2NaOH -> Na2SO4 + 2H2O 2. We need moles of H2SO4 to find the concentration. So we can find first the moles of NaOH and use a mole:mole ratio to find the moles of H2SO4. *Don't forget, change all mL > L. Use formula Molarity= moles/L and thus moles=M x L 0.01314L x 0.35M= 0.00460 mol NaOH 0.00460 mol NaOH x (1 mol H2SO4/ 2 mol NaOH)= 0.0023 mol H2SO4 3. Now we can find the molarity. 0.0023 mol / 0.01802L = 0.13 M H2SO4

Calculate the molarity of Na+ ions in 135 mL of a solution containing 5.32 g NaCl and 1.20 g Na2SO4. A.0.737 B.0.706 C.0.0995 D.0.799 E.0.108

D. 0.799 convert grams > moles of each compound. You should have around 0.091 mols of Na from NaCl, and 0.017 mols of Na from Na2SO4. Work: (5.32g NaCl) x (1mol/58.44g NaCl) x (1 mol Na/1 mol NaCl)= 0.091 (1.20g Na2SO4) x (1 mol/142.04) x (2 mol Na/1 mol Na2SO4)= 0.017 0.019 + 0.017= 0.108 mol Na Molarity= moles/ Liters 0.108 mol Na/ .135 L= 0.8M or 0.799 M

Balance the following redox equation using the smallest integers possible and select the correct coefficient for the hydrogen sulfite ion, HSO3¯.MnO4¯(aq) + HSO3¯(aq) + H+(aq) → Mn2+(aq) + SO4^2¯(aq) + H2O(l) A.1 B.2 C.3 D.5 E.10

D. 5

Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O(l) ΔH°f = -1411 kJ C(s) + O2(g)--> CO2(g) ΔH°f = -393.5 kJ H2(g) + 1/2 O2(g)--> H2O(l) ΔH°f = -285.8 kJ A.731 kJ B.87 kJ C.1.41 x 103 kJ D.52 kJ E.2.77 x 103 kJ

D. 52 kJ ΔH°f = 2 * (-393.5) + 2 * (-285.8) - 1 * (-1411) = +52.4 kJ/mol

Which of the following compounds is a weak electrolyte? A.LiNO3 B.H2SO4 C.MgI2 D.HF E.CoSO4

D. HF A. Nitrate- always soluble B. SO4- soluble (some exceptions) C. Group 1 metals= soluble D. Not soluble, but not listed as unsoluble= weak electrolyte E. soluble

Which set contains only the FALSE statement(s)in the group below? a. For the freezing of liquid water ΔH is positive. b.For a general exothermic reaction ΔH is positive. c. For the combustion of 1 mol of methane in oxygen ΔH is negative. A.a. only B.b. only C.c. only D.a. and b. E.a., b., and c.

D. a. and b a. ΔH is negative for water freezing because a change in state from a liquid to solid, it releases heat even though its cold, thus making in exothermic. b. exothermic will have a -ΔH. c. combustion= release heat= -ΔH

The oxidation state of vanadium in [HV10O28]5- is: A.-6 B.+8 C.-5 D.+3 E.+5

E. +5

Calculate the work for the expansion of CO2 from 1.0 to 3.0 liters against a pressure of 1.0 atm at constant temperature. A.3.0 L·atm B.0 L·atm C.-3.0 L·atm D.2.0 L·atm E.-2.0 L·atm

E. -2.0 L·atm w=-PΔV w=-(1)(3-1)= -2 L atm

Calculate the molarity of a 23.55-mL solution which contains 28.24 mg of sodium sulfate (м = 139.04 g/mol). A.8.625 M B.1.199 M C.0.8339 M D.0.2031 M E.0.008625 M

E. 0.008625 M

A gas evolved during the fermentation of alcohol was collected at 17°C and 746 mm Hg. After purification its volume was found to be 19.4 L. How many moles of gas were collected? A.203 moles B.10.5 moles C.608 moles D.13.6 moles E.0.80 moles

E. 0.80 moles n=PV/RT Convert mm Hg to atm. 746 mm Hg x (1 atm/ 760 mm Hg)= 0.98 atm Plug it all in -> (0.98atm x 19.4L)/ (0.082 x 290K) = 0.79 = 0.80 mol

If the barometer read 765.2 mmHg when the measurement in in the Figure below took place, what is the pressure of the gas in the flask in kilopascals? *Picture shows a manometer with Mercury higher on the left side, so the Pgas < Patmosphere. There is a change in height= 1.30 cm A.7.55 kPa B.102.4 kPa C.1.007 kPa D.752.2 kPa E.100.3 kPa

E. 100.3 kPa When the pressure is greater on the gas (flask) side of the manometer, where Pgas > Patmosphere, we would add Patm + h. This would be the opposite in this case where Pgas < Patmosphere, so we would subtract Patm- h. Before we start plugging in #'s check the units! First we must know to convert to kPa at the end because we have 765.2 mmHg, then we need the height to be the same as mmHg so we'll first convert it to mm. 1.30 cm x (10 mm/ 1 cm)= 13 mm Now plug it all in and convert to kPa. 765.2 mm Hg - 13 mm= 752 mmHg x (1 atm/ 760 mm Hg) x (101.325 kPa / 1 atm) = 100.2 kPa

After balancing the following redox equation, the sum of the stoichiometric coefficients on the right hand side of the equation is:Cu^2+(aq) + Ce^3+(aq) + H2O => CeO2(s) + Cu(s) + H^+(aq) A.2 B.3 C.7 D.9 E.11

E. 11 1. Determine the oxidation #'s and which one is being oxidized and reduced. Here Cu is being reduced (+2 > 0) and Ce is being oxidized (+3 > +4). 2. Separate them into half reactions. Cu 2+ -> Cu Ce 3+ + H2O-> CeO2 + H 3. Balance elements with a coefficient (other than O and H). 4. Add H2O to balance O, and H to balance H. Ce 3+ + 2H2O -> CeO2 + 4H 5. Add e- to balance charge. Cu 2+ + 2e- -> Cu We'll add 2e- to the left to have them both have a 0 charge. Ce 3+ + 2H2O -> CeO2 + 4H + 1e- (3+0=+3) vs (0+4=+4) so we'll add 1 e- to the right side so they're both +3. 6. Multiply half reactions so the number of e- is equal, so we'll multiply our reaction with Ce by 2. (Ce 3+ + 2H2O -> CeO2 + 2H + 1e-) x2 -> 2Ce 3+ + 4H2O -> 2CeO2 + 8H + 2e- 7. Combine the 2 and cancel anything that is the same on both sides. Cu 2+ + 2e- 2Ce 3+ + 4H2O -> 2CeO2 + 8H + 2e- + Cu Here we'll cancel out 2e-. Cu 2+ 2Ce 3+ + 4H2O -> 2CeO2 + 8H + Cu Add up the right side> 2 + 8 + 1= 11

A mixture of oxygen gas, nitrogen gas, and carbon dioxide in a 10.0 L container at 25 oC has a total pressure of 12.5 atm. If there are 30.0 g each of oxygen and nitrogen, how many grams of carbon dioxide are present? A.5.12 g CO2 B.1.10 g CO2 C.48.3 g CO2 D.3.11 g CO2 E.137 g CO2

E. 137 g CO2 Working backwards... We need grams of CO2, so we need mols of CO2 to find it. Recall Dalton's law, where their mole fractions all= 1. If we can find the mole fraction of the other 2 then we can find the mols of CO2. So for mole fractions we need the mols of each (by converting grams>mols) and total mols (by using n=PV/RT). 1. Total mols=n=(12.5 x 10)/(0.082 x 298)= 5.11 total mols 2. Mols of O2= 30g x (1 mol/ 32g)= 0.94 mol O2 Mols of N2= 30g x ( mol/ 28g)= 1.07 mols N2 3. Solve for mole fraction Mole fraction of O2= 0.94/5.11= 0.18 Mole fraction of N2= 1.07/ 5.11= 0.21 4. All of the mole fractions will always equal 1. So we'll subtract them from 1 to find the mole fraction of CO2. 1-0.18-0.21=0.61 mole fraction of CO2. 5. Solve for mols using mole fraction formula. 0.61= (x mols of CO2/ 5.11 mols)= 3.12 mols of CO2 6. Convert to grams 3.12 mol x (44g CO2/ 1 mol)= 137g CO2

What is the density of Freon-11 (CFCl3) at 120°C and 1.5 atm? A.1.92 g/L B.52.3 g/L C.5.23 g/L D.0.192 g/L E.6.39 g/L

E. 6.39 g/L We want the density but we don't have the mass, but we do have mols and volume where we can convert mols to grams. But what is the value? PV=nrt where n/V=P/RT to get moles/vol. (1.5)/(0.082)(393)= 0.046 mol/L CFCl3=136 g/mol 136 g/mol x (0.046 mol/L)= 6.39 g/L

A 3.0-L sample of helium was placed in a container fitted with a porous membrane. Half of the helium effused through the membrane in 24 hours. A 3.0-L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse through the membrane? A.8.5 h B.12 h C.48 h D.60. h E.68 h

E. 68 h (R1/R2)= √(M2/M1) Rate of effusion is in the units of L(volume) or moles per unit of time. So we can determine the rate of effusion for He. (1.5 L/24 h)= 0.0625 L/h. Determine the Molar masses and solve for R2. (0.0625/R2)=√(32/4)= 0.022 L/h Now we can convert 1.5 h to the hours it takes for O2 to diffuse. (1.5L) x (1 hr/ 0.022 L)= 68 h

According to the first law of thermodynamics, the energy of the universe is constant. Does this mean that ΔE is always equal to zero? A.No, ΔE never equals zero because energy is always flowing between the system and surroundings. B.Yes, ΔE = 0 at all times, which is why q = -w. C.No, ΔE never equals zero because work is always being done on the system or by the system. D.No, ΔE does not always equal zero, but this is only due to factors like friction and heat. E.No, ΔE does not always equal zero because it refers to the system's internal energy, which is affected by heat and work.

E. No, ΔE does not always equal zero because it refers to the system's internal energy, which is affected by heat and work.

In a reaction, a reducing agent A.gains electrons. B.causes the oxidation of another compound. C.is reduced in a reaction. D.is a spectator ion. E.lowers the oxidation number of an atom in another compound.

E. lowers the oxidation number of an atom in another compound.

A 25.0 g sample of garden compost was analyzed for chloride content. The sample was dissolved in water and the chloride was precipitated as silver chloride. 1.58 g of dried precipitate was obtained. Calculate the percent chloride in the sample. A.6.3% B.1.37% C.62% D.2.4% E.1.56%

E. 1.56% Determine the formula: AgCl. We need grams of Cl first to find the percent of it in the sample. We'll Convent: Grams of AgCl > Mols of Cl > Grams of Cl> % in sample (1.58 g AgCl) x ( 1 mol/ 143g AgCl) x (1 mol Cl/1 mol AgCl)= 0.011 mol Cl 0.011 mol Cl x (35 g Cl/1 mol)= .385 g Cl We have a proportions of (.385 g Cl/25 g)=(x/100%) (.385g x 100/25g)= x = 1.54% *It's a little different due to how I rounded masses.

The pressure a gas would exert under ideal conditions is always greater than the observed pressure of a real gas. True False

False A gas behaves more like an ideal gas at higher temperature and lower pressure