book even answer ch3-4

book even answer ch3-4 - 4. 16. 30. 34. 1.00 mol CaC12 x...

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Unformatted text preview: 4. 16. 30. 34. 1.00 mol CaC12 x 1.00 mol CaC12 x CHAPTER 3—4 CHAPTER 3 6.022 x 1023 formula units mol 6.022 x 1023 formula units mol X X 1 Ca2+ formula unit 2 Cl' formula unit a. solution — a homogeneous mixture of two or more substances b. solvent - the substance that other substances are dissolved in c. solute- the substance being dissolved d. molarity - the molar amount of solute per liter of solution e. dilute solution - a relatively small amount of solute in a given amount of = 6.02 x 1023 Ca2+ = 1.20 x 1024 Cl' f. concentrated solution - a relatively large amount of solute in a given amount of solution solution a. 3.16 x 102416115 b. 3.82 x 1023 molecules c. 5.68 x 10-22 g a. c 3 x 12.01 = 36.03 H 8 x 1.008 = 8.064 0 1 x 16.00 = 16.00 60.09 2% 0 £313: x 100% = 59.96% c 7% H = 26933: x 100% = 13.42% H _ 16.00 g _ o .% o — 60.09 g x 100% — 26.63%, 0 b. N 2 x 14.01 = 28.02 H 8 x 1.008 = 8.064 s 1 x 32.07 = 32.07 0 4 x 16.00 = 64.00 132.15 _ 28.02 g o _ o ?%N — 13215 g x 100A) — 21.20%, N _ 8.064 g o _ o 7% H — 13215 g x 100A) — 6.102%. H _ 32.07 g o _ o 7% s — 13215 g x 100% — 24.27%. s _ 64.00 g o = o 7% 0 — 13215 g x 100/o 4843/0 o c. C 6 x 12.01 = 72.06 H 8 x 1.008 = 8.064 N 2 x 14.01 = 28.02 1 m n. mm M,” .11. amnmpumvlri‘ «v- pump-mm 01 w uw‘rvmmwfisn“ u » m mm ‘1 m 1'11mnfnmnmmflm’Vn'” ., ,0 44. 48. 108.14 72. ?% c = 108012: x 100% = 66.64% C ?% H = 18626112: x 100% = 7.457% H 28.02 ?% N =108 14‘: x 100% = 25.91% N d. Ba 1 x 137.3 = 137.33 C1 2 x 35.45 = 70.90 0 11 x 16.00 = 176.00 H 6 x 1.008 = 6.048 390.28 9% Ba = g x 100% = 35.19% Ba ?% c1= 379238: x 100% = 18.17% c1 7% o = 3332: x100% = 45.10% 0 12% H = 3693421: x 100% = 1.550% H molecular formula: Na2S405 . 3 x 18 W03 3H2° %H2°-7Tl4—+mm 6 x 18 . 0 _—_ _— MgClz “120 A H20 24 + 71 + (6 x 18) ZnSO4-7H20 % H20 = 7 x 18 65+32+64+(7x18) 9x18 “(No3)3'9H20 % H20 = 27 + 42 +48 + 48 +"48 + (9 x 18) Estimates give the values Li 69325321320) ~16299 ~<°5 Mg 953621320) ~ g5) ~>°'5 Zn fing”;—3g”<05 A1 _______—9"2° ~33 ~<0.5 70 +150 + (9 x 20) 400 Actual 44% 53% 44% 43% 52. 58. 64. 68. 74. 78. 84. MgC12'6H20 is the only one that is greater than 50%. It is the largest % H20. CH4N a. 2 KC103(s) -——> 2 KCl(s) + 3 02(g) IOH b. CH3CHCH2CH3(1) + 6 02(g)-> 4 C02(g) + 5 H20(l c- 4 NH3(g) + 5 02(g) —> 4 N0(g) + 6 H20(g) d. C12(g) + 2 NH3(g) + 2 NaOH(aq) __> 2 H20(£’) + 2 NaCl(aq) + N2H4(aq) a. 932 g HNO3 b. 2.04 x 103 g C7H5N306 26.1 L CH3CH20H H20 is limiting and can produce 0.313 mol HBF4_ 4NO+02+2Na2CO3—>4NaN02+2C02 9 N _ 1 4 mol Na2CO3 4 molNaN02 3' ‘ g aNOZ ‘ 5 g Naco3 X 106.0 g Na2CO3 X 2 mol Na2CO3 X 69.00 g NaN02 _ m — 20° 3 NaN02 m0] N0 4 mol NaN02 69.00 g NaN02 9 = _———- —— __ 'g NaN02 ‘05 g NO X 30.01 gNO X 4 molNO X mol NaN02 = 241 g NaN02 mol ()2 4 mol NaN02 69.00 g NaN02 9 = — _______._.. —_ ' g NaN02 75'0 g 02 X 32.00 g 02 X m0102 mol NaN02 = 647 g NaN02 There are 200 g of NaN02 produced. b. NO and 02 are in excess. mol NaN02 4 mol NO 30.01 g NO ? g NO ‘ 200 g NaN02 X 69.00 g NaN02 X 4 mol NaN02 X mol N0 = 87.0 g NO used ? g N0 excess = 105 g - 87 g =18 g NO excess 9 mol NaN02 m0102 32.00 g 02 - g 02 ‘ 200 g NaN02 X 69.00 g NaN02 X 4 mol NaN02 X mol 02 = 23.2 g 02 used ? g 02 excess = 75.0 g - 23.2 g = 51.8 g 02 excess. a 0.500 M H2SO4 b. 5.71 M C2H50H c. 6.324 M KOH d. 0.0364 M H2C204 e. 0.07770 M C6H14O4 f. 0.783 M C3H9N 94. a. 96.03 mL b. 21.0 mL 98. 1.33 M 100. 32.8 mL HCl Chapter 4 2. a. salt b. strong base c. salt d. weak acid e. strong acid f. weak base g. salt h. strong base 4. (d) [ions] = 0.075 M 10. Ca(OH)2 (s) + 2 H+ (aq) —> Ca2+ (aq) + 2 H20 (1) 14. ZnC12 + MgSO4 will not react. The possible products (ZnSO4 and MgClz) are soluble. Zn(OH)2 will precipitate. 2112+ +,e’1' +,K+ + OH'—y Zn(OH)2(s) +,K+ + 31- Zn2+ + OH' —> Zn(OH)2(s) 16. Na2CO3. Most carbonates are insoluble. c032- + Mg2+ —> MgCO3(s) 3o. [Li+] = 0.015 + (2 x 0.015) = 0.045 M, [01-] = 0.015 + (3 x 0.015) = 0.060 M, [Mg2+] = 0.015 M, [r] = 2 x 0.015 = 0.030 M, [3042-] = 0.015 M, [Al-3+] = 0.015 M H20 40. a. HNO3 (1) — > H+(aq)+N03-(aq) b. CH3(CH2)2COOH(aq) <=> H+(aq) + CH3(CH2)2COO'(aq) H20 c. Ba(OH)2 (s)— > Ba2+ (aq) + 20H' (aq) d. HC102 (aq) <=> H+ (aq) + C102' (aq) e. HC204' (aq) <=> H+ (aq) + cz042- (aq) f. (CH3)2NH(aq) + H20 <=> (CH3)2NH§(aq) + 011-00 44. HCO3' (aq) + HCOOH (aq) —-> HCOO' (aq) + H2CO3 |—> H20 + C02 (g) 48. 4.027 M 56. a. Ba2++ c032- —> BaCO3 (s) b. sz+ + s042- —> PbSO4 (s) c. No reaction d. 0H' + HSO4' —> H20 (1) + $042- 2 OH' + Mg2+ —-> Mg(OH)2 (s) CH3CH2COOH + OH' —> H20 (1) + CH3CH2COO‘ Ba2+ (aq) + s2- (aq) + Cu2+ (aq) + 3042- (aq) —> BaSO4 (s) + CuS (s) . am» 0) + 3 H100 —> Cr3+ (aq) + 3 H20 (1) NH3 (aq) +H+ (aq) —> NH4+ (aq) . No reaction e. 2 OH‘ (aq) + Mg2+ —> Mg(OH)2 (s) 58. 9.00-er9 62. 2 Fe3+ (aq) + 3 32- (aq) —> Fe283 (s) 66. a. neither. The oxidation states do not change. b. neither. The oxidation states do not change. 0. reduction. The oxidation state of carbon decreases from 2 to -4. ...
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book even answer ch3-4 - 4. 16. 30. 34. 1.00 mol CaC12 x...

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