For all phase changes the sign of Δ S and Δ H is always going to be the sameSublimation, melting, vaporization = +Freezing, deposition, condensation = -Correct answer is ΔS > 0

Vapor pressure increases as IMF’s get weakerRanking IMF:Ionic strongest H-bond dipole dipole London Dispersion ForcesWhat does each have?BeO – IonicEthyl Amine( CH3CH3NH2) - H-BondingBenzene - (C6H6) CsF- ionicC3H6O (acetone)- dipole-dipoleN2- LDFWe have two ionic salts so you rank them based on charge densityBeO has much large charge density so therefore will have a lower vapor pressureCorrect answer is #2

So if the ΔH lattice is larger ( always a positive number) than ΔH hydration (always a negative number) then which is true?So if ΔH lattice is a large positive number and ΔH solution is a smaller negative number than ΔH solution is going to be positive and endothermicSo salt can dissolve if temperature of the solution is increased.

Triple point is intersection of three phases. This can be solid liquid gas or three different phases it doesn’t matterSo the triple point here is at 200K and 50 atm. Answer choice #2

So container is at 150 atm and at 25 degrees C. Convert to K which is a classic mistake and thus we are at 300K and 150 atm which is in the Liquid area of the graph which you will have to label. So we are at star one and when the container is opened the pressure is relieved (violently). The liquid would vaporize (something like a propane tank).

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Question type is : calculating delta H from heating across phase transitions.

So we have ice steam and water all togetherWe need to bring them all in the same phase to figure our how we can find what the equilibrium temp is. First lets see how much energy it takes to melt the 80 g of ice and get 160 g of water at 0 degrees Cq=m ΔH fusion (80g) (334 J/g) =26720 joules of heatNow we have 160 grams of water at 0 degrees C on one side of the equationOn the other side we have 16 g of steam. Lets take this 26720 joules of heat and use it to cool down the 120 degree steam to hot water at 100 degrees CSo first we cool down the steam to 100 degrees C steam. q=m Csteam ΔT (16g)(2.08 J /g C) ( 20 C) 665.6 JNow lets turn this 100 degrees C steam to 100 degrees C waterq=m ΔH vap (100g) (2256 J/g) = 36160 JNow on the cooling down steam side we have 36160 + 665.6 J = 36825.6 JOn the water side we have 26720 joules. Therefore if you take the difference between the water and steam you have 36825.6 J – 26720 J= 10105.6 J of heat that will heat up the water.q=m Cwater ΔT = 10105.6 = (160g)(4.184 J/gC)(X) X =15.09 degrees CFinally we have the same phase now we just have to do an equilibrium problem. We have on one side 160 g water at 15.09 degrees C and on the other we have 16g water at 100 degrees C m Cwater ΔT = m Cwater ΔT (160 g)(4.184 J/g C)(T eq - 15.09 )=(16g)(4.184 J/gC)(T eq - 100)Solve for T eq = 176 grams water at 22.8 degrees C

Solubility of gas is exothermic therefore it is inversely proportional to temperature. Also solubility of gas is directly proportional to pressure. N2 is a non-polar molecule and there fore a less polar solvent would help solubilitySo the answer is I and III

Like dissolves like so octane which is non-polar will dissolve in non-polar mediaTherefore the most polar will be least soluable for gasolineOil/ fuel are going to be non-polarWax is also non-polarSugar is polar with hydrogen bondsSalt is the most polar –ionicSo the answer is salt

Temperature and pressure has exponential relationship

#2 statement would raise the boiling point

#3-#4 are correct

We need to figure out molar massSo pressure= Moles/liter * R *tempLets turn 1500 Pa to atm 1 atm= 101325 pa1500 pa = .014 atm So .014 atm = X/.2L *0.0821 * 300 KSolve for x = .000114 molesMoles= Grams/molar mass.000114= 1/X X is molar massX= 8771.93 choice 2

We want grams of salti =2 for NaClMolality = moles /kg solventT= 44= 2 * X/2 *.512 X=7.81We have 7.81 moles of NaCl To get grams we multiply by molar mass of NaCl which is = 58.44456 grams so answer #4

Increase , entropy

Reactants over productsUse solutions and gassesTherefore the correct answer is #4

Kc= [H2][I2]/[HI]2

[H2] at eq is .233 thus [I2] must also be .233Kc =.012Rearrange equation and solve for [HI]

[HI]2 =[H2][I2]/Kc

[HI] =([H2][I2]/Kc) ^(1/2)

[HI] = [(.233)(.233)/(.12)] ^(1/2) answer choice 5

Kc= 4Initial concs areNO = 5 initial Change -X eq 5-X CO2 = 10 initial change -X eq 10-XNO2 = 2 initial change+X eq 2+XCO = 0 initial change +X X

Kc= 11.7 SO2, O2, SO3 = .015Kc= [SO3] 2/[SO2]2 [O2]

(.015)^2/(.015)^2 (.015)1/.015 = 66.666So Q > K reaction will shift to the leftSo answer choice is #3

Le Chatelier says if we add product reaction will shift to the left to reestablish eq

So this is an endothermic reactionSo as T goes up K will go upSo answer choice one is correct

If K is a very small number or less than 1 delta G will be a positive numberIf K is a large number greater or than 1 delta G will be negativeIf K is 1 then delta G is 0So answer choice 5 is impossible

It will never be O

Kw gets larger as T increases… so answer I is wrongKw is 10 ^-14 at room temp this is correct[H+]=Kw/[OH-] so answer choice 3 is incorrect

So only true answer is II

[H+]=Kw/[OH-]

1.7x10^-7= 1x10^-14/XSolve for XX = 6.7X10^-8 Answer choice 4

BaF2 [Ba2+] 2[F-]

Ksp = X * (2X)2

Ksp = X * 4X2

Ksp = 4X3

2x10^-6 = 4X3

Solve for XX= .0079 mol Answer choice 4

Just rank them based on KspLargest Ksp will be most solubleThus the correct order is #3

Weakest base here will be the strongest acid…. And the strongest acid will have the largest KaD- will therefore be the weakest Base

Ca=[H+] for strong acids Cb=[OH-] for strong bases

Weak acids and bases[H+]= (CaKa)^(1/2)[OH-]= (CbKb)^(1/2)

[H+]= (CaKa)^(1/2)[H+]= [(0.10)(1X10^-6)]^(1/2)[H+]= 3X10^-4-log(3X10^-4) = 3.5 Answer #5

Weak acids and bases[H+]= (CaKa)^(1/2)[OH-]= (CbKb)^(1/2)

We have NaF which gives us F- in solution which is the conjugate base of HFSo we need to get the Kb from HF’s KaKw/Ka=Kb1X10^-14/7.2X10^-4)= kbKb= 1.3x10^-11

[OH-]= (CbKb)^(1/2)[H+]= [(0.30)(1.3X10^-11)]^(1/2)[OH-]= 2 X 10^-6-log(2X10^-6) =5.6 which is our pOHSo pH = 14- 5.6= 8.3 Answer #1

This is a buffer calc because we have B and BH+

Kb=1.8X10^-5 pKb= 4.744.74 + log(.3/.1)= pOH4.74 +.38 =pOH5.12= pOH10^-5.12= [OH-]7.5X10^-6 = [OH-]