Exam 1 Fall 11 CHM 2045 final version Form A w out ans
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Exam 1 Fall 11 CHM 2045 final version Form A w out ans

Course Number: CHEM 2045, Fall 2011

College/University: University of Florida

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CHM 2045 Exam 1 (Gower, Mitchell, Williams) (Form Code A) Fall 2011 Instructions: This exam contains 25 questions worth 10 points each. On your Scantron form enter and bubble in your name, UF ID number (start on the leftmost space and leave any extra spaces blank), and Form Code (see above). You may retain your exam sheet (mark your answers on it and the Scantron) and any other sheets. Turn in only your...

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2045 Exam CHM 1 (Gower, Mitchell, Williams) (Form Code A) Fall 2011 Instructions: This exam contains 25 questions worth 10 points each. On your Scantron form enter and bubble in your name, UF ID number (start on the leftmost space and leave any extra spaces blank), and Form Code (see above). You may retain your exam sheet (mark your answers on it and the Scantron) and any other sheets. Turn in only your Scantron. Check your bubbling carefully minus 9 points for bubbling errors, non-negotiable. 1. If you mix 250. mL 1.0 M KCl with 375. mL 2.5 M KCl, what is the molarity of KCl in the final solution? (1) 0.4 M (2) 1.9 M (3) 1.2 M (4) 1.6 M (5) 2.4 M 2. Which choice(s) contain(s) both ionic and covalent bonds? I. H3O+ II. Mg(OH)2 III. SF6 IV. Na2O V. K2CO3 (1) I only (2) I & II only (3) I & III only (4) II & V only (5) I & V only 3. Which choice is correctly classified ? (1) chromium, main group element (2) ammonia, ionic compound (4) hydride, polyatomic anion (5) butane, homogeneous mixture (3) xenon, monatomic gas 4. A mixture of copper(II) sulfate pentahydrate (249.70 g/mol) and magnesium sulfate heptahydrate (246.49 g/mol) is heated until all the water is lost. The weight of the initial mixture is 5.020 g, and the weight of the resulting anydrous compounds is 2.988 g. Which choice is closest to the percent by mass of copper(II) sulfate pentahydrate in the original mixture? (1) 40.48% (2) 50.32% (3) 59.52% (4) 53.54% (5) 70.86% 5. A 500. mL sample of 2.00 M HCl solution is treated with 4.47 g of magnesium. Calculate the concentration of the acid solution after all the metal has reacted. Assume that the volume remains unchanged. (The products of this reaction are magnesium chloride and hydrogen gas.) (1) 1.63 M (2) 1.26 M (3) 0.632 M (4) 0.368 M (5) 0.184 M 6. Consider the reaction MnO2(s) + 4HCl(aq) MnCl2(aq) + Cl2(g) + 2H2O(l). If 74.8 g MnO2 and 48.2 g HCl react, what is the theoretical yield of Cl2 closest to? (1) 23.4 g (2) 61.0 g (3) 93.7 g (4) 11.7 g (5) 46.9 g 7. What volume of a 0.500 M hydrochloric acid solution is needed to neutralize each of the following, respectively: a) 10.00 mL 0.300 M sodium hydroxide ; b) 10.00 mL 0.200 M barium hydroxide ? (1) 6.00 mL; 4.00 mL (2) 3.00 mL ; 4.00 mL (3) 6.00 mL ; 8.00 mL (4) 6.00 mL ; 2.00 mL (5) 1.50 mL ; 2.00 mL 8. An iron ore sample contains Fe2O3 plus other non-iron impurities. A 752 g sample of the ore is heated with excess carbon, producing 253 g of pure iron by the following reaction: Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g) What is the mass percent of Fe2O3 in the original iron ore sample? Assume that the reaction is 100% efficient. (1) 48.1% (2) 96.2% (3) 33.6% (4) 19.24% (5) 16.8% 9. Excess ice at 0C is placed in an insulated cup containing 361 g of soft drink at 23C. The specific heat of the drink is the same as that of water (4.184 J/gC). Some ice remains after the drink cools to 0C. Determine the mass of ice that has melted. Assume that the heat capacity of the cup can be ignored. It takes 334 J to melt 1.00 g of ice at 0C. (1) 52 g (2) 104 g (3) 123 g (4) 204 g (5) 275 g 10. Which pair of atoms below has the same number of neutrons? 64 68 64 63 28 Ni , 29 Cu , 30 Zn , 30 Zn ? (1) 63 29 Cu & 64 30 Zn (b) 64 28 Ni & 63 29 Cu (c) 64 28 Ni & 64 30 Zn (d) 63 29 Cu & 68 30 Zn (e) 64 30 Zn & 68 30 Zn 11. Methanol (CH3OH) can be manufactured by the reaction of gaseous carbon monoxide and hydrogen. Suppose 68.5 kg of gaseous carbon monoxide is reacted with 8.60 kg of gaseous hydrogen. If 37.5 kg of methanol is actually produced, what is the percent yield of the reaction? (1) 51.5% (2) 47.9% (3) 27.4% (4) 48.6% (5) 54.9% 12. All of the following are named correctly EXCEPT (1) LiClO4; lithium (2) perchlorate CaHPO4; calcium hydrogen phosphide (3) NaCN; sodium cyanide (4) Mg(OH)2; magnesium hydroxide (5) CaSO3; calcium sulfite (Form Code A) 13. The following enthalpy data are given: Reaction H (kJ/mol) -1225.6 I. P4(s) + 6Cl2(g) 4PCl3(g) -2967.3 II. P4(s) + 5O2(g) P4O10(g) -84.2 III. PCl3(g) + Cl2(g) PCl5(g) -285.7 IV. PCl3(g) + (1/2)O2(g) Cl3PO(g) What is H (kJ/mol) for the reaction P4O10(g) + 6PCl5(g) 10Cl3PO(g) (1) -110.5 (2) -610.1 (3) -2682.2 (4) -7555.0 (5) +3186.8 14. Which statement(s) describe(s) the difference between the charge of a polyatomic ion and the oxidation states of its constituent atoms? I. The charge of a polyatomic ion is a property of the entire ion, while the oxidation states are assigned to each individual atom. II. The oxidation state of the polyatomic ion is the same as the charge. III. The charge of a polyatomic ion is not a real physical property, while the oxidation states of atoms are actual physical properties. (1) I only (2) I & II only (3) I & III only (4) I, II & III (5) None of these statements is correct. 15. The compound adrenaline contains 56.79% C, 6.56% H, 28.37% O, and 8.28% N by mass. What is the empirical formula for adrenaline ? (1) C9H17ON2 (2) C6H5O4N2 (3) C8H13O2N2 (4) C8H11O3N (5) C9H17ON2 16. What is the total number of ions in 6.36 g of SrF2 ? (1) 1.83 x 1023 ions (2) 6.09 x 1022 ions (3) 3.65 x 1023 ions (4) 9.15 x 1022 ions (5) 5.48 x 1023 ions 17. Which compound is matched with the correct oxidation number for nitrogen ? (1) NH3, zero (2) N2O5, +3 (3) NaNO2, +3 (4) NH2OH, +1 (5) N2O, +2 18. A 90.0 mL sample of 0.640 M NaBr is mixed with 220. mL of a SrBr2 solution. If the final solution has a bromide ion concentration of 1.26 M, what was the concentration of the SrBr2 solution ? (Assume that volumes are additive.) (1) 1.52 M (2) 0.950 M (3) 1.20 M (4) 0.601 M (5) 0.757 M 19. What mass of sodium iodide is needed to react exactly with 250. mL 0.0113 M lead(II) nitrate? (1) 0.847 g (2) 0.423 g (3) 0.00565 g (4) 1.69 g (5) 3.75 g 20. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l), H = 1.37 103 kJ/mol For the reaction above, which of the following statements is(are) true? I. The reaction is exothermic. II. The enthalpy change would be different if gaseous water were produced. III. The reaction is not an oxidationreduction reaction. (1) I & III only (2) I, II, & III (3) II & III only (4) I & II only (5) I only 21. A +1 ion contains 80 electrons and 124 neutrons. Which choice is an isotope of this ion? (1) 205Tl+ (2) 202Hg+ (3) 205Hg+ (4) 203Tl+ (5) 206Pb+ 22. Five beakers contain 100 mL samples of the following 5 different 0.1 M solutions (different solution in each beaker). A 100mL aliquot of 0.1 M magnesium bromide is added to each beaker. Which beaker forms a precipitate? (1) 0.1 M CuSO4 (2) 0.1 M AgNO3 (3) 0.1 M K2SO4 (4) 0.1 M NH4Cl (5) 0.1 M Mn(NO3)2 23. A 0.2254 g sample of naphthalene (C10H8, 128.16 g/mol) was burned in a bomb calorimeter with a heat capacity of 2.18 kJ/K, and the temperature increased by 4.16C. What is the molar energy for combustion for this compound? (1) 5.16x103 kJ/mol (2) 6.51x105 kJ/mol (3) 22.5 kJ/mol (4) 5.07 kJ/mol (5) 8.91 J/mol 24. Which choice is a correctly written formula for an ionic compound? (1) ZnCl (2) Li2SO4 (3) Tl+NO3 (4) (NH4)Br (5) Cu2S2 25. On Earth, there is only one stable isotope of sodium, 23Na with an isotopic mass of 22.98977 amu. However, on mythical planet Mirth there is another stable isotope, 25Na with an isotopic mass of 24.94638 amu. Analysis of samples taken from various locations on Mirth give an average result of 24.62 amu for the atomic mass of sodium. What is the percent abundance of 25Na on Mirth? (1) 85.26% (2) 50.25% (3) 75.26% (4) 83.32% (5) 70.23%

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University of Florida - CHEM - 2045
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University of Florida - CHEM - 2045
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University of Florida - CHEM - 2045
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UNF - MAP - 2302
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UNF - MAP - 2302
31.71.81.91.101.111.12One implementation for the sparse matrix is described in Section 12.3 Another implementation is a hash table whose search key is a concatenation of the matrix coordinates.Every problem certainly does not have an algorithm. As
UNF - MAP - 2302
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UNF - MAP - 2302
2Mathematical Preliminaries2.1(a) Not reexive if the set has any members. One could argue it is symmetric, antisymmetric, and transitive, since no element violate any ofthe rules.(b) Not reexive (for any female). Not symmetric (consider a brother and
UNF - MAP - 2302
6Chap. 2 Mathematical Preliminaries(d) This is not an equivalance relation since it is not symmetric. For example, a = 1 and b = 2.(e) This is an eqivalance relation that divides the rationals based on theirfractional values. It is reexive since for a
UNF - MAP - 2302
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UNF - MAP - 2302
8Chap. 2 Mathematical Preliminaries2.13 void allpermute(int array[], int n, int currpos) cfw_if (currpos = (n-1) cfw_printout(array);return;for (int i=currpos; i<n; i+) cfw_swap(array, currpos, i);allpermute(array, n, currpos+1);swap(array, curr
UNF - MAP - 2302
9p2q 2 = 4( )22p22q = 2( )2This implies that q 2 is also even. Thus, p and q are both even, which contradicts the requirement that p and q have no common factors. Thus, 2 mustbe irrational.2.17 The leftmost summation sums the integers from 1 to
UNF - MAP - 2302
10Chap. 2 Mathematical Preliminaries(c) Induction Step.ni=112in1=i=1= 111+ni2212n1+12n1.2nThus, the theorem is proved by mathematical induction.2.20 Proof:(a) Base case. For n = 0, 20 = 21 1 = 1. Thus, the formula is correctfor
UNF - MAP - 2302
112.22 Theorem 2.1ni=1 (2i)= n2 + n.(a) Proof: We know from Example 2.3 that the sum of the rst n oddnumbers is n2 . The ith even number is simply one greater than the ithodd number. Since we are adding n such numbers, the sum must be ngreater, or
UNF - MAP - 2302
12Chap. 2 Mathematical Preliminaries8 5 n2()3355< ( )2 ( )n2335n=.3=Thus, the theorem is proved by mathematical induction.2.24 Proof:22(a) Base case. For n = 1, 13 = 1 (1+1) = 1. Thus, the formula is correct4for the base case.(b) Induc
UNF - MAP - 2302
13iii. Induction Step. Consider the case where n + 1 pigeons are in nholes. Eliminate one hole at random. If it contains one pigeon,eliminate it as well, and by the induction hypothesis some otherhole must contain at least two pigeons. If it contains
UNF - MAP - 2302
14Chap. 2 Mathematical PreliminariesInduction hypothesis: T(n 1) = 2n1 1.Induction step:T(n) = 2T(n 1) + 1= 2(2n1 1) + 1= 2n 1.Thus, as proved by mathematical induction, this formula is indeed the correctclosed form solution for the recurrence.2.
UNF - MAP - 2302
152.32 (I saw this problem in John Bentleys Programming Pearls.) Approach 1:The model is Depth X Width X Flow where Depth and Width are in milesand Flow is in miles/day. The Mississippi river at its mouth is about 1/4 milewide and 100 feet (1/50 mile)
UNF - MAP - 2302
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UNF - MAP - 2302
3Algorithm Analysis3.1 Note that n is a positive integer.5n log n is most efcient for n = 1.2n is most efcient when 2 n 4.10n is most efcient for all n > 5. 20n and 2n are never moreefcient than the other choices.3.2Both log3 n and log2 n will hav
UNF - MAP - 2302
18Chap. 3 Algorithm Analysis3.8 Other values for n0 and c are possible than what is given here.(a) The upper bound is O(n) for n0 > 0 and c = c1 . The lower bound is(n) for n0 > 0 and c = c1 .(b) The upper bound is O(n3 ) for n0 > c3 and c = c2 + 1.
UNF - MAP - 2302
193.11(a)nnn! = n (n 1) ( 1) 2 122nnn 1 1 1222n n/2=()2Thereforenlg n! lg2(b) This part is easy, since clearlyn21 (n lg n n).21 2 3 n < n n n n,so n! < nn yielding log n! < n log n.3.12 Clearly this recurrence is in O(log n n)
UNF - MAP - 2302
20Chap. 3 Algorithm Analysis3.15 Yes. When we specify an upper or lower bound, that merely states our knowledge of the situation. If they do not meet, that merely means that we dontKNOW more about the problem. When we understand the problem completely,
UNF - MAP - 2302
21binary search of the subarray will nd the position n in an additional log nsearches at most, for a total cost in O(log n) searches.3.19 Here is a description for a simple (n2 ) algorithm.boolean Corner(int n, int m, Piece P1, Piece* array) cfw_for
UNF - MAP - 2302
22Chap. 3 Algorithm AnalysisSWAP(array[i][j], array[tempr][tempc]);Finding the corner takes O(n2 m2 ) time, which is the square of the number ofpieces. Filling in the rest of the pieces also takes O(n2 m2 ) time, the numberof pieces squared. Thus,
UNF - MAP - 2302
4Lists, Stacks, and Queues4.1 Call the list in question L1.L1.setStart();L1.next();L1.next();val = L1.remove();4.2(a) | 10, 20, 15 .(b) 39 | 12, 10, 20, 15 .4.3 list L1(20);L1.append(2);L1.append(23);L1.append(15);L1.append(5);L1.append(9);
UNF - MAP - 2302
24Chap. 4 Lists, Stacks, and Queueslink<Elem>* temp1 = head->next;link<Elem>* temp2 = temp1->next;while (temp2 != NULL) cfw_link<Elem>* temp3 = temp2->next;temp2->next = temp1;temp1 = temp2;temp2 = temp3;head->next = temp1;4.6(a) The followin
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25/ Move curr to prev positiontemplate <class Elem>void LList<Elem>:prev() cfw_link* temp = curr;while (temp->next!=curr) temp=temp->next;curr = temp;(b) The answer is rather similar to that of Part (a).4.7 The space required by the array-based l
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26Chap. 4 Lists, Stacks, and Queues4.10 I assume an int requires 4 bytes, a double requires 8 bytes, and a pointerrequires 4 bytes.(a) Since E = 4 and P = 4, the break-even point occurs when1n = 4D/8 = D.2Thus, the linked list is more space efcien