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Final Jeopardy Question Thermo- Chemistry ( H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Page 1: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.
Page 2: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

Final Jeopardy QuestionFinal Jeopardy Question

Thermo-Chemistry

(H)Solutions

500

chemical

Reactions

BondingPart II

Liquids

&

Solids

Gases

100

200

300

400

500 500 500 500

400 400 400 400

300 300 300300

200 200 200 200

100 100 100100 100

200

300

400

500

Page 3: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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You dissolve 65 grams Li2S in 195 mL of 31°C water. After dissolving, the temperature of the solution is 20°C. A) what is Hsoln Li2S, including the correct unit and sign, +/-, and B) what is the system and what is the surroundings, and C) is the process exothermic or endothermic?

Hwater = (195 g)(-11°C)(4.18 J/g°C) = -8966.1 J This is the energy change of this specific process, but Hsoln is the energy change per mol, so divide the energy change by the amount of moles:

+8966.1 J = +6331 J/mol = Hsoln Li2S 65 g /45.9 g/mol

Note that the sign changes due to it being endothermic from the system point of view.

Why do you suppose Hsoln is expressed per mol?

Page 4: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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rxn progress

1

2

3

4

5 Label each of 1-5;explain what is occurring in 2,explain what 4 and 5 are.

1) Energy of reactants, 2) energy released by thereaction, H, 3) energy of products, 4) energyof activation, EA: energy required for the reactionto occur, 5) activation complex: single particleformed by colliding reactants existingmomentarily.

Page 5: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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For the reaction

If Hrxn = -0.85 kJ, 1) What mass of magnesium will produce 6500 joules of energy?2) Rewrite the reaction with the energy term as part of the reaction3) Is the reaction endothermic or exothermic?

Mg(s) + Au2(SO4)3(aq) MgSO4(aq) + Au(s)

3Mg(s) + Au2(SO4)3(aq) 3MgSO4(aq) + 2Au(s) + 0.85 kJ

24.3 g

Page 6: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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You add 0 °C ice to 25 cm3 of 48 °C water.What mass of ice will cool the water to 41 °C?

Additional info: Cliquid H2O = 4.18 J/g °C, Cice = 2.1 J/g °C, Hfus = +6.01 kJ/mol

Hice = Hwater

(X g ice/18 g/mol)(6010 J/mol) = (25 g)(7 °C)( 4.18 J/g °C)

X = 2.19 g ice

Page 7: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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You run out of water during your summer trip to Isla Grevy. It’s -38 °C outside.You break off some surface ice and your scale tells you it is 2480 grams.What amount of heat energy will you need to warm it up to 30 °C?

ADDITIONAL INFO: Cliquid H2O = 4.18 J/g °C, Cice = 2.1 J/g °C, Hfus = +6.01 kJ/mol

Answer = 1,343,967 J = 1344 kJ (38)(2480)(2.1)+ (2480/18)(6010)+ (30)(2480)(4.18)

0

-38

30

T1

T2

H1

H2 H3

Page 8: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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You have a solution of zinc chloride andyou need to precipitate out the cation.What compound could you use?Write the complete balanced reactionwith states included.

Any soluble compound that contains any of the following anions:carbonate, hydroxide, phosphate, sulfide, sulfite,As long as its cation does NOT precipitate chloride—then youwould have two precipitates!

Page 9: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Which of the following will dissolve ina non-polar solvent? Explain your reasoning.

a) CHCl3 b) BCl3 c)NF3 d) HF

a, c, and d are polar, b is non-polar: a is symmetrical but has different outside atoms;c is not symmetrical and so the polarity of its bonds do not cancel;d is not symmetrical and so the polarity of its bond does not cancel;b is trigonal planar, and therefore the polarity of the B-Cl bonds cancel, so it is non-polar and will therefore dissolve in a non-polar solvent.

Page 10: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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What is Tfp when dissolving 55 g sodium sulfite in 555 cm3 of water?

for Na2SO3, i = 3

T = imKfp m = molality = mols solute/kg solvent

Page 11: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

1) If the solution is on the line, then the maximumamount of solute is dissolved, so it is saturated but there is no undissolved solute. Above the line, it is still saturated, but becausethere is more than the maximum amount, there is undissolved solute, and a dynamic equilibrium is established: dissolved solute joins the undissolved solute, while undissolved solute becomes solvated. This occurs at the same rate, and so the amount of dissolved and undissolved remains constant.2) Solubility at 30°C = 82 g/200 g H2O. Solubility at 50°C = 102 g/200 g H2O.102 – 82 = 20 grams must be added to keep it saturatedat the higher temperature. Back

1) What is the difference between being onthe solubility line and above the line?Assume no supersaturation has occurred!2) You have 200 mL of a saturated solutionof ammonium chloride at 30 °C. What mass of solute must be added to saturate it if the temperature is raised to 50 °C?

0 10 20 30 40 50 60 70 80 90 100

SOLUBILITY CURVES150

140

130

120

110

100

90

80

70

60

50

40

30

20

10

0

TEMPERATURE oC

KNO3

SO2

Page 12: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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In general, what is the minimum size ofparticles considered to be a colloid?

It must be > 1 nm, which is large enough toreflect light, which creates a Tyndall beam.

Page 13: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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What makes the slope of water’ssolid-liquid equilibrium line interesting?

Lots ‘o stuff….

Page 14: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Draw two kinetic energy distributioncurves of two samples of a substanceat two different temperatures.Assume that each sample has thesame amount of particles. Label whichis warmer, which is colder

A and B represent the two averagekinetic energies. B > A, so the red temperature isgreater than the blue temperature.Same amount of particles in bothmeans the area under each curve must be equal.

Page 15: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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1) At normal pressure and roomtemp, what would happen to solidsolid iodine if the temperature waswas raised by 50 °C?

2) Under what conditions could you expect iodine to exist as a liquid?

phase diagram of iodine

Normal P = 1 atm, room temp = ~25 °C. A 50 °C increase would be ~75 °C.At 25 °C and 1 atm, I2 is a solid, at 75 °C I2 is a gas, so it will sublimate.Iodine exists as a liquid under conditions of T and P indicated by the greenishshaded area on the graph.

Page 16: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Vapor pressure of water

Temperature Pressure (̊#C) (̊mmHg)

10.0 9.2 20.0 17.5 30.0 31.8 40.0 55.3 50.0 92.5 60.0 149.4 70.0 233.7 80.0 355.1 90.0 525.8 100.0 760.0

The table at right gives various vapor pressuresfor water at various temperatures.The vapor pressure increases with temperature.Would you expect this to be the case with any liquid?Explain.

As can be seen by the graph from the answer toLiquids & Solids-400 pts, as KE increases, more andMore particles have enough energy to break awayfrom the liquid and become a gas. Since it is thevapor (gas) particles that create vapor pressure,the more vapor given off by the liquid, the higherthe vapor pressure will be.

Page 17: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Liquid nitrogen (bp = -194 °C) is used by physicians to kill unwanted cells on the skin (warts, pre-cancerous tissue, etc.) by freezing the cells. Why would freezing a cell kill it? Would you expect it to just thaw back to its pre-frozen state?

Since water expands when it freezes, the expanded water (ice) will burstthe cell, thus killing it.

Page 18: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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What is the hybridization of nitrogen’s valenceelectrons in the molecule at right?Draw using arrows and boxes,include the name(s) of the orbital(s);AND name the shape at the nitrogen and at the carbon!

H—C=N—H

H

Nitrogen has 5 valence e—. It has three bonds, one pi andtwo sigma, and one pair of unbonded electrons. Electronsin pi bonds come from unhybridized p-electrons, so one of nitrogen’s p valence e— remains unhybridized so it canpi-bond to carbon. The rest of the e— are hybridized as shown above.SHAPES: Since N has one pair of unbonded electrons, that pushes the bondsout of line and it is therefore bent. Carbon has three electron groups around it,all bonded, so they spread out in a trigonal planar shape.

Page 19: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Water has a very low molar mass;Why is it a liquid at room temperature

rather than a gas?

Of the two molecules pictured below, C2H6 (ethane) and H2CO (formaldehyde),which would need more kinetic energy to break away from its liquid state andbecome a gas? (both are 30 g/mol) Explain your answer.

The shape and bond polarities of formaldehyde give it a strong dipole.Ethane is completely symmetrical and non-polar. The stronger dipole-dipole interactions of formaldehyde would require greater energy tobreak away from the liquid.

Page 20: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Is the compound at rightpolar or non-polar?Explain your reasoning!

Are there any unbonded electron pairs on this central atom?Explain how you know.(hint: each bond is a single bond)What is its VSEPR shape?Would you expect it to be a polar molecule?

Without an unbonded pair on the central atom, theshape would be trigonal planar. Here, the bonds have been pushed out of the plane,so there must be an unbonded pair repelling the bonds.The shape is pyramidal.Pyramidal is not symmetrical, and so IF the bonds are polar, then the molecule would bepolar as well—the bond polarities would not cancel due to the non-symmetrical shape:the polarity vectors of the bonds would not add up to zero.

Page 21: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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How many -bonds are in C3H4?Show the dot and dash structure—the carbons are in the middle.

C=C=CH

H

H

H

There are two double bonds, each has one pi and one sigma,so there are a total of two pi bonds.

Page 22: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Why do atoms form bonds?

Water has the lowest mass of anymolecule except hydrogen;

Why is it a liquid at room temperaturerather than a gas?

Is the compound at right polar or non-polar?Explain your reasoning!

The molecule is tetrahedral, which is a symmetrical shape.All outside atoms are the same, so any bond polarity would be cancelleddue to the symmetry: the bond polarity vectors would add up to zero.The molecule is therefore non-polar.

Page 23: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

358

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For the combustion of 90 grams of ethene, C2H4,What volume of carbon dioxide is produced at 690K and 103 kPa?

C2H4 + 3O2 2CO2 + 2H2O

Page 24: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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3 g of a gas in a 400 mL sealed container are held at 151 K and 1.62 atm. What is the molar mass of the gas?

Page 25: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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25 grams C2H4 is at 300 kPa in a volume of 2500 mL. What is it’s temperature?

Page 26: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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A balloon originally at 500 torr, 30 °C, and 35 mL,changes to 400 torr and 40 mL.What would the temperature be under the new conditions?

277 K

Page 27: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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The two balloons at rightare at the same T and P,their volume ratio is 5:9.What relationship amongthe four gas variablesgives you that volume ratio?

Volume, V, is directly proportional to the amount of particles, n,when T and P are held constant.

Page 28: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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What is the molar mass of Pb3P4?

What is the mass of 6.02 x 1023 particles of Pb3P4?

What is the mass of 1 mole of Pb3P4?

What is the mass of 1 mole of Pb3P4 formula units?For the reaction of aluminum with aqueous copper II nitrate, what is

A)The oxidation 2Al 2Al3+ + 6 e-

B)The reduction 3Cu2+ + 6 e- 3Cu

C) The oxidizing agent Cu2+

D) The reducing agent AlE)The number of electrons exchanged in the balanced equation 6

2Al + 3Cu(NO3)2 2Al(NO3)3 + 3Cu

Page 29: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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For the reaction ZnF2(aq) + AlCl3(aq) What minimum mass of ZnF2 solid must be dissolved in 555 mL of 0.666 M AlCl3 to react completely?

3ZnF2(aq) + 2AlCl3(aq) 3ZnCl2(aq) + 2AlF3(s)

Page 30: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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In a combustion reaction,22 g ethene, C2H4, reacts with 52 grams oxygen.

Which is limiting?

C2H4 + 3O2 2CO2 + 2H2O

needed

75.4 g oxygen is needed to react with 22 g ethene, butthere are only 52 g oxygen available, so oxygen is limiting.

have

Page 31: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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For the reactionMg(s) + H2O(l) Mg(OH)2(aq) + H2(g)

what volume of hydrogen gas is producedfrom the reaction of 500 grams water

in excess magnesium at STP?

Page 32: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

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Write the net ionic equation for the following reaction

K2CO3(aq) + ThCl4(aq) KCl(aq) + Th(CO3)2(s)

Ionic: 4K+ + 2CO3

2- + Th4+ + 4Cl- 2K+ + 4Cl- + Th(CO3)2(s)

Net ionic: 2CO32- + Th4+ Th(CO3)2(s)

Page 33: Final Jeopardy Question Thermo- Chemistry (  H) Solutions 500 c hemical R eactions B onding Part II L iquids & S olids G ases 100 200 300 400 500 400.

What is the basisfor

Avogadro’s Number?

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What two units used in chemistryare derived from carbon-12,and how are they derived?