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ACHM 120 Review Exam 2

Chapters 4, 5 & 6
by

Ashley Twible

on 15 October 2012

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Transcript of ACHM 120 Review Exam 2

Chapters 4, 5 & 6 EXAM 2 Review Chapter 4 Chapter 5 Chapter 6 Problem 1 Key Equations
Chapter 4 Molarity = moles solute volume sol'n (L) _______________ Dilution = M V x M V init. init. dil. dil. Predict whether each of the following compounds is soluble in water: a) AgI
b) Na CO
c) BaCl
d) Al(OH)
e) Zn(Ch COO) 2 3 2 3 2 Answer 1 Solubility chart - Table 4.1 on page 121 a) insoluble
b) soluble
c) soluble
d) insoluble
e) soluble Problem 2 Identify the precipitate (if any) that forms when the following solutions are mixed, and write the balanced equation for each reaction. a) NaCH COO and HCl
b) KOH and Cu(NO )
c) Na S and CdSO 3 3 2 2 4 Answer 2 a) NaCH COO + HCl ---> CH COOH + NaCl 3 (aq) (aq) (aq) (aq) 3 No Precipitate formed; all products soluble b) 2 KOH + Cu(NO ) ---> 2 KNO + Cu(OH) 3 2 3 2 (aq) (aq) (aq) (s) c) Na S + CdSO ---> CdS + Na SO 2 4 2 4 (aq) (aq) (aq) (s) Problem 3 Write balanced net ionic equations for the reactions that occur in each of the following cases: a) Cr (SO ) + (NH ) CO ---> 2 4 3 (aq) 4 3 2 (aq) b) Ba(NO ) + K SO ---> 3 2 (aq) 2 4 (aq) c) Fe(NO ) + KOH ---> 3 2 (aq) (aq) Answer 3 a) 2 Cr + 3 CO ---> Cr (CO ) 3+ (aq) 3 (aq) 2- 2 3 3 (s) Spectators: NH , SO 4 + 4 2- b) Ba + SO ---> BaSO 2+ (aq) 4 2- (aq) (s) Spectators: K , NO + 3 - c) Fe + 2 OH ---> Fe(OH) 2+ (aq) - (aq) 2 (s) Spectators: K , NO + - 3 Problem 4 Which of the following solutions is the most basic? a) 0.6M NH
b) 0.150M KOH
c) 0,100M Ba(OH) 3 2 Answer 4 The two strong bases of the three are KOH and BA(OH) 2 Ba(OH) has twice as many basic components per molecule 2 Therefore, 0.100M of Ba(OH) is the strongest base 2 Problem 5 An aqueous solution of an unknown solute is tested with litmus paper and found to be acidic. The solution is weakly conducting compared with a solution of NaCl of the same concentration. Which of the following substances could be the unknown? a) KOH
b) NH
c) HNO
d) KClO
e) H PO
f) CH COCH 3 3 2 3 3 3 3 Answer 5 We know that the solution conducts some electricity so it must be an electrolyte. It conducts less than a know strong electrolyte, NaCl. Therefore, it must be a weak electrolyte. The weak electrolytes from the options are NH and H PO . 3 3 3 Since the problem tells us the solution is acidic, only one of the two weak electrolytes is an acid. Therefore our answer is H PO . 3 3 Problem 6 Write the balanced molecular and net ionic equations for each of the following neutralization reactions: a) aqueous acetic acid is neutralized by barium hydroxide
b) solid chromium(III) hydroxide reacts with nitrous acid
c) aqueous nitric acid and aqueous ammonia react Answer 6 a) 2 CH COOH + Ba(OH) ---> Ba(CH COO) + 2 H O 3 2 3 2 2 (l) (aq) (aq) (aq) Net Ionic = CH COOH + OH ---> CH COO + H O 3 (aq) - (aq) (aq) 2 (l) b) Cr(OH) + 3 HNO ---> Cr(NO ) + 3 H O 3 (aq) (aq) 2 2 3 (aq) 2 (l) Net Ionic = Cr(OH) + 3 HNO ---> 3 H O + Cr + 3 NO 3 (s) 2 (aq) 2 (l) 3+ (aq) 2 (aq) - c) HNO + NH ---> NH NO 3 (aq) 3 (aq) 4 3 (aq) Net Ionic = H + NH ---> NH + (aq) 3 (aq) 4 + (aq) Problem 7 Determine the oxidation number for the indicated element in each of the following compounds: a) Co in LiCoO
b) Al in NaAlH
c) C in CH OH
d) N in GaN
e) Cl in HClO
f) Cr in BaCrO 2 4 3 2 4 Answer 7 a) 3+ d) 3-
b) 3+ e) 3+
c) 2- f) 6+ Problem 8 Which of the following are redox reactions? For those that are, indicate the element that is oxidized and reduced. a) P + 10 HClO + 6 H O ---> 4 H PO + 10 HCl 4 (s) (aq) 2 (l) 3 4 (aq) (aq) b) Br + 2 K ---> 2 KBr 2 (l) (s) (s) c) CH CH OH + 3 O ---> 3 H O + 2 CO 3 2 (l) 2 (g) 2 (l) 2 (g) d) ZnCl + 2 NaOH ---> Zn(OH) + 2 NaCl 2 (aq) (aq) 2 (s) (aq) Answer 8 a) redox reaction ; P is oxidized and Cl is reduced

b) redox reaction ; K is oxidized and Br is reduced

c) redox reaction ; C is oxidized and O is reduced

d) precipitation reaction Problem 9 a) Calculate the molarity of a solution made by dissolving 12.5 grams of Na CrO in enough water to form exactly 550mL of solution.

b) How many moles of KBr are present in 150mL of a 0.275M solution?

c) How many milliliters of 6.1M HCl solution are needed to obtain 0.100 mol of HCl? Answer 9 a) M = 12.5g Na CrO 2 4 _____________ 0.550 L x 1 mol Na CrO 2 4 __________________ 161.97g Na CrO 2 4 = 0.140M Na CrO 2 4 b) moles = 0.275M KBr x 0.150L = 4.13 x 10 mol KBr -2 c) 6.1M = 0.100 mol HCl _____________ L L = 0.016 which = 16mL Problem 10 a) How many milliliters of a stock solution of 6.0M HNO would you have to use to prepare 110mL of 0.500M HNO ? b) If you dilute 10.0mL of the stock solution to a final volume of 0.250L, what would the concentration of the diluted solution be? Answer 10 a) (6.0M)V = (0.500M)(110mL) 1 V = 9.2mL HNO 1 3 b) (6.0M)(10.0mL) = (250mL)M 2 M = 0.24M HNO 2 3 3 3 Problem 1 Key Equations
Chapter 5 E = mv K 1 2 __ 2 = Kinetic Energy deltaE = E - E = q + w final initial Change in Internal Energy deltaH = deltaE + P(deltaV) Enthalpy change at constant pressure q = C x m x deltaT s = Heat gained/lost deltaH = sum(deltaH) - sum(deltaH) products reactants Indicate which of the following is independent of the path by which change occurs: a) the change in potential energy when a book is transferred from table to shelf

b) the heat evolved when a cube of sugar is oxidized to CO

c) the work accomplished in burning a gallon of gasoline 2 (g) Answer 1 a) Potential Energy is a state function - independent b) heat is not a state function - dependent



c) work is not a state function - dependent Problem 2 At one time, a common means of forming small quantities of qxygen gas in the laboratory was to heat KClO : 3 2 KClO ---> 2 KCl + 3 O deltaH = -89.4kJ 3 (s) (s) 2 (g) For this reaction, calculate deltaH for the formation of: a) 1.36mol of O
b) 10.4g of KCl
c) The decomp. of KClO proceeds spontaneously when heated. Explain what conditions the reverse reaction is possible. 2 3 Answer 2 a) 1.36mol O 2 ( ________________ ) -89.4 kJ 3mol O 2 = -40.5 kJ b) 10.4g KCl ( ) ( ) ________________ _________________ 1mol KCl 74.55g KCl -89.4 kJ 2mol KCl = -6.24 kJ c) Since the sign of the reverse reaction is opposite, is seems reasonable that other characteristics would be reversed as well. The reverse reaction is not spontaneous and requires energy to proceed. Problem 3 When a 4.25g sample of solid ammonium nitrate dissolves in 60.0g of water in a coffee-cup calorimeter, the temperature drops from 22.0 C to 16.9 C. Calculate deltaH (in kJ/mol NH NO ) for the solution process: 4 3 0 0 NH NO ---> NH + NO 4 3 (s) 4 + (aq) 3 - (aq) Assume that the specific heat of the solution is the same as that of pure water (4.18 J/g- C). Is this process endothermic or exothermic? Answer 3 Total mass = 60.0g H O + 4.25g NH NO = 64.3g 2 4 3 deltaT = 22 C - 16.9 C = 5.1 C 0 0 0 q = (0.00418 kJ/g- C)(64.3g)(5.1 C) = 1.371 = 1.4kJ 0 0 0 This is the heat absorbed when NH NO dissolves 4 3 Thus, when 4.25g NH NO , the ratio of heat per mole of NH NO is 4 3 4 3 1.4kJ ( _____________ 4.25g NH NO 4 3 1mol NH NO 4 3 ) ( _____________________ 80.04g NH NO 4 3 1mol NH NO 4 3 ) = 25.82 kJ/mol NH NO
or 26kJ/mol 4 3 Problem 4 From the enthalpies of reaction: 2 C + O ---> 2 CO deltaH = -221.0 kJ (s) 2 (g) (g) 2 C + O + 4 H ---> 2 CH OH deltaH = -402.4 kJ (s) 2 (g) 2 (g) 3 (g) Calculate deltaH for the reaction: CO + 2 H ---> CH OH (g) 2 (g) 3 (g) Answer 4 You first want to flip the first reaction so that all your target reactanct and products in the reaction you are trying to get to are all on the correct sides. After rewritting the resulting reaction and deltaH it should be apparent that you need to divide the entire reaction by 2 in order to get the desired molar amounts. The resulting deltaH = (-402.4kJ+221kJ=-181.4kJ)/2 = -90.7 kJ
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