| 1-34 |
(a) 6.32 x 1025 (b) 7.82 x 10-17 (c) 0.2020 (d) 0.01161 (e) 2.29 x 10-27 (the subtraction has to be done first, and that causes loss of s.f.'s) (f) 0.81 (g) 1.130 x 103 |
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| 2-54 |
(a) Al3+ (b) Se2-
(c) Ba2+ (d) N3-
(e) Fr+ (f) Br- |
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| 2-56 |
(a) mercury(I) oxide (b) iron(III) bromide (c) cobalt(II) sulfide (d) titanium(IV) chloride (e) Sn3N2 (f) CoI3 (g) HgO (h) CrS3 |
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| 3-104 |
We must find the limiting reagent: (PhCl = chlorobenzene,
MM = 112.55 g/mol) (CL = chloral, MM = 147.38 g/mol) (DDT, MM = 354.46 g/mol) nPhCl = 1142 g / 112.55 g/mol = 10.15 mol nCL = 485 g / 147.38 g/mol = 3.29 mol (b) Because there is a 2:1 molar ratio, every 1 mol of CL will consume 2 mol of PhCl. Thus, 3.29 mol of CL requires 6.58 mol of PhCl. Therefore, PhCl is in excess and CL is the limiting reagent. (a) mDDT = (nCL)*(1 mol DDT / 1 mol CL)*(MM DDT) = 3.29 mol * 354.46 g/mol = 1170 mol DDT formed. (c) moles PhCl left over = 10.15 mol - 6.58 mol = 3.57 mol mass left over = 3.57 mol * 112.55 g/mol = 402 g (d) theoretical yield was 1170, actual yield was 200.0 g % yield = 200.0/1170 * 100% = 17.1% |
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| 3-118 |
cabon disulfide is CS2, hydrogen sulfide
is H2S (a) CH4 (g) + 4 S (s) ---> CS2 (g) + 2 H2S (g) (b) There is 120. g / 16.05 g/mol = 7.48 mol of methane There is 120. g / 32.07 g/mol = 3.74 mol sulfur 7.48 mol of methane would require 29.92 mol of sulfur. Since there is much less sulfur here, it is the limiting reagent. Therefore, mass CS2 = (3.74 mol S) (1 mol CS2 / 4 mol S) (76.15 g/mol) = 71.2 g CS2. |
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| 4-30 |
1.00 L of 0.50 M requires (1.00 L * 0.50 mol/l
= ) 0.50 mol of substance. (a) 0.50 mol / 18 mol/L = 28 mL (b) 0.50 mol / 12 mol/L = 42 mL (c) remember that 1 mol NiCl = 1 mol of NiCl.6H2O (MM = 237.69 g/mol) 0.50 mol * 237.69 g/mol = 120 g (d) 0.50 mol / 16 mol/L = 31 mL (e) sodium carbonate = Na2CO3 (MM = 105.99 g/mol) 0.50 mol *105.99 g/mol = 53 g.. |
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| 4-34 |
1.584 g of Mn becomes 1.584 g / (54.94 g/mol) = 0.02883
mol M of stock solution = 0.02833 mol / 1.000 L = 0.02833 mol/L Solution A: 50.00 mL * 0.02833 mol/L / 1000.0 mL = 0.001442 mol/L Solution B: 10.00 mL * 0.001442 mol/L / 250.0 mL = 5.766 x 10-5 mol/L Solution C: 10.00 mL * 5.766 x 10-5 mol/L / 500.0 mL = 1.153 x 10-6 mol/L |
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| 4-58 |
(a) full equation: AgOH (s) + HBr
(aq) ---> H2O (l) + AgBr
(s) full ionic: AgOH (s) + H+ (aq) + Br- (aq) ---> H2O (l) + AgBr (s) net ionic: AgOH (s) + H+ (aq) + Br- (aq) ---> H2O (l) + AgBr (s) (b) full: Sr(OH)2 (aq) + 2 HI (aq) ---> 2 H2O (l) + SrI2 (aq) full ionic: Sr2+ (aq) + 2 OH- (aq) + 2 H+ (aq) + 2 I- (aq) ---> 2 H2O (l) + Sr2+ (aq) + 2 I- (aq) net ionic: OH- (aq) + H+ (aq) ---> 2 H2O (l) (c) full: Cr(OH)3 (s) + 3 HNO3 (aq) ---> 3 H2O (l) + Cr(NO3)3 (aq) full ionic: Cr(OH)3 (s) + 3 H+ (aq) + 3 NO3- (aq) ---> 3 H2O (l) + Cr3+ (aq) + 3 NO3- (aq) net ionic: Cr(OH)3 (s) + 3 H+ (aq) ---> 3 H2O (l) + Cr3+ (aq) |
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| 4-62 |
Amount of H+ from HCl: (0.0500 L) (0.100
mol/L) = 0.00500 mol Amount of H+ from HNO3: (0.1000 L) (0.200 mol/L) = 0.0200 mol Total H+: 0.0250 mol Amount of OH- from Ca(OH)2: (0.5000 L) (2 mol OH- / 1 mol Ca(OH)2) (0.0100 mol/L) = 0.0100 mol Amount of OH- from RbOH: (0.200 L) (0.100 mol/L) = 0.0200 mol Total OH-: 0.0300 mol Therefore, there is an excess of 0.0300 - 0.0250 = 0.0050 mol of OH- |
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Last updated, September 27, 2006
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