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CH9_Page_12

CH9_Page_12 - P351 =10“ 26.4x I'll" bar=ﬂ.ti4n1bar...

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Unformatted text preview: P351 =10“: 26.4x I'll" bar=ﬂ.ti4n1bar pm =1.4x10‘3bar—2{s.4xlethal-Floats“ bar=ﬂ.12mbar The percentage decomposition is given by 2 (0.4 x 1040a) 100:010’ Islam—Jam x n 9.51} (a) concentration of PC15 initiallyr = 2.0 g Pet5 203.22 gﬂ'ﬂDFI P'UlS 0.300 L = {1.032 mol - L'L Concentration [mol‘L'lju PCl5(g] # PC13[g) + (312(3) initial 0.032 0 0 change —x +1: +3: ﬁnal 9.032 - x +3: +3: K: = [PC13][C]2] = [x}[x) = x2 [PCIS] (0.032 — x] (0.032 — x} x2 — .m. 0.01 (0.032 — x] x2 = (0.01)t0.032 — x) x2 + (asljx — 0.020 = 0 = —(IJ.61) a 100.01}? — (amt—0.020) 2-1 I _ {0.61) i 0.6? 2 -l x = +0.03 or -ﬂ.ﬁ4 The negative root is not meaningful, so we choose it = 0.03 mol - L". [PCIJ] = [C12] = 0.03 mol+ L"; [PCIS] = {1032 — 013-3 mol - L"i = (lﬂtlﬂ mo]. L'l. The solution of this problem again points up the problems with following the signiﬁcant ﬁgure conventions for calculations of this type, because we would use onlyr one signiﬁcant ﬁgure as imposed by the subtraction in the quadratic equation 263 ...
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