3600_Ch3_probs - 3-1 (a) 3-2. (a) 3-3. (a) 3-5. (a) (6)...

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Unformatted text preview: 3-1 (a) 3-2. (a) 3-3. (a) 3-5. (a) (6) 3-6. (a) (b) 3-7. (a) (6) 35-11. (a) (b) (0) 3-12. 3-13. (a) Carmen 5 (b) 4 (C) 3 1.237 (b) 1.238 (c) 0.135 (d) 2.1 (e) 2.00 0.217 (b) 0.216 (C) 0.217 3.71 (b) 10.7 (c) 4.0><101 (d) 2.85x10'6 12.6251 (f) 6.0><10'4 (g) 242 BaClg = 137.327 + 2(35.452 7) = 208.232 C6H4O4 = 602.0107) + 40.007 94) + 405.999 4) = 140.093 5 (The 4th decimal place in the atomic mass of C has an uncertainty of i8 and the 4th decimal place of O has an uncertainty of i 3. These uncertainties are large enough to make the 4th decimal place in the molecular mass of C6H404 insignificant. The best answer is 140.094.) 12.3 (b) 75.5 3.04x10'10 0) 11.9 (c) 5.520x 103 (d) 3.04 (g) 4.600 (h) 4.9><10‘7 25.031 mL is a systematic error. The pipet always delivers more than it is rated for. The number i 0.009 is the random error in the volume delivered. The volume fluctuates around 25.031 by i 0.009 mL. The numbers 1.98 and 2.03 mL are systematic errors. The buret delivers too little between 0 and 2 mL and too much between 2 and 4 mL. The observed variations i 0.01 and i 0.02 are random errors. The difference between 1.9839 and 1.9900 g is random error. The mass will probably be different the next time I try the same procedure. (b) Cynthia (c) Chastity ((1) Cheryl 3.124 (i0.005), 3.124 (i0.2%). It would also be reasonable to keep an additional digit : 3.1236 (i0.0052), 3.1236 (i0.17%) 3-14. 3-22. (a) 6.2(i0.2) — 4.1 (i0.1) 2.1 i e e2 = 022+ 0.12 => 6 = 0.22 Answer: 2.1i 0.2 (or 2.1: 11%) (b) 9.43 (:005) x 0.016 (:0001) 9.43 (i0.53%) => x 0.016 (i6.25%1 %62 = 0.532 + 6.252 0.150 88 (i %e) :> %e = 6.272 Relative uncertainty = 6.27%; Absolute uncertainty = 0.150 88 x 0.062 7 = 0.009 46; Answer: 0.151 i 0.009 (or 0.151 i 6%) (c) The first term in brackets is the same as part (a), so we can rewrite the problem as 2.1 (i0.224) + 9.43 (i005) = 2.1 (i10.7%) + 9.43 (i0.53%) %e = 'V10.72 +0.532 = 10.7% Absolute uncertainty = 0.107 x 0.223 = 0.023 9 Answer: 0.223 i0.024 (ill%) (d) The term in brackets is 6.2 (i0.2) x 10'3 + 4.1 (i0.1) x 10'3 10.3 (i0.224) x 10'3 = 10.3 x 10'3 (i2.17%) 9.43 (i0.53%) x 0.010 3 (i2.17%) = 0.097 13 i2.26% = 0.09713 $0.002 20 0.0971 : 00023 (i 2.3%) e = V0.22+ 0.12 => 6 = 0.224 Answer: mol H+ = 2 x mol NagCOg 0.9674 (i0.000 9) g 0.9674 (i0.093%) g molNa2C03 = 105.988 (220.001) g/mol = 105.988 (50.000 94%) g/mol = 0.0091274 (:0.093%) mol molH+ = 2(0.0091274(i0.093%)) = 0.018255 (i0.093%)mol (Relative error is not affected by the multiplication by 2 because both mol H+ and uncertainty in mol H+ are both multiplied by 2.) 1 . fHCl _ 0.018255 (i0.093%) mol _ 0.018255 (i0.093%)mol m0 my" - 0.027 35 @0000 04) L ' 0.027 35 (i0.146%) L = 0.66746 (i0.173%) = 0.667 46 (i0.001 155) = 0.667 i 0.001 M ...
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3600_Ch3_probs - 3-1 (a) 3-2. (a) 3-3. (a) 3-5. (a) (6)...

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