1、解：_X定量化學(xué)分析簡明教程習(xí)題解答第二章20.48 +20.55 +20.58 +20.60 +20.53 +20.50(%) = 20.54%6中位數(shù)：20.54%極差：R=(20.60-20.48)%=0.12%i 訣別為：-0.06, 0.01,0.04,0.06,-0.01,-0.04(%)0.060.010.04 - 0.06 -= = - % = 0.037 %0.040.060.010.042 2 2 060.010.04- % = 0.046 %6 -1s0.046%X 200.046%0.22%= 0.019%（2） 絕關(guān)于誤差 Ea=20.54%-20.45%=0.09%

2、相關(guān)于誤差 Er=0.09% -20.45%=0.44%解：（1）35.72%35.60%為可疑值Q 35.60%35.72%的取舍35 .60 % -35 .65 %Q計算35 .72 % -35 .60 %0 .42查表90%, n = 4時，Q=表. 35 .60 %保留35 .72 % -35 .69 %Q計算35 .72 % -35 .60 %0.25查表置信度為90%,n=4時，Q=0.76,. 35.72%保留表-35.72 +35.69 +35.65X4% = 35 .66 %s = 0.052 %(2)tsr - 90% 時,二：-35 n2.350.052%二 35 .66

3、 % _ 0.06%卜=95%時,=35 .66%3.180.052%二 35.66 % - 0.08 %1（X CF - ,XaU,naaU所以 4 次測定時的置信區(qū)間為0.12%0.12%9.56%-9 .56 % .1.9644解：解法（1）x = 16 .72 %s=0.08%16.72%0.08%x = 2.5 c t.95(3)= 3.18解法 (2)ts=x 二:-16.72% 3.18 ”0.08 % = (16.72 0.13)%解:.n真值落在此區(qū)間內(nèi)無系統(tǒng)誤差XX=96.4%,=93.9%Xi2S=0.9%22大F 計：大0.9 2=9.12.與SpS0.6 2顯著性差異

4、、2 (4 1)4(5 520.90.8X Xng96.4 -93.9i2-4.6 -t= 2.37(0.95 7)兩組數(shù)據(jù)平均值有顯著性差異，.有系統(tǒng)誤差，即溫度關(guān)于測定結(jié)果有影響。2解:0.20520.2042Q計算0.20520.20380.71Q：=0 .76表第三個結(jié)果不應(yīng)舍去結(jié)果應(yīng)表示：=0.2043S=0.2052 -0.2042IQ計算=-=0.71.Q=0.64表第三個結(jié)果應(yīng)舍去解：2HCI+N&CO=2NaCI+CO2+HOm(Na2CO)=(1/2) c(HCl) V(HCl)M(Na2C( )31 1=(1/2)0.1mol1 10.025L 106gmo=0.13gE

5、 = _0.2%r2HCl+ NazBO 1OHO = 4MBO+2NaCI+5HOm (NazBO 1OHO)=(1/2) c(HCl) V(HCl) M(Na2BQ1OHO)=(1/2)-10.1mol L 0.02-L 381g mol =0.48g從稱量誤差考慮稱取NaO10HO能達到_0.1%的準確度，2B解：0.0030(二), 6.023 1023(四),64.120(五),4.80 10-10(三),998(三),10000(無法確定),1.0 103(二)pH5. 2,H+=6 10-5 mol L (pH=5.2 為一位有效數(shù)字)-1解：(1)213.64+4.4+0.3244=218.3(2)0.0982 X(20 .00 -14 .39) X162 .206/31.4182 X10001002.10 %二(3)pH12. 20,H+=6.3 10-13 mol L (pH=12.20 為兩位有效數(shù)字)-1解：c(ETA) V ( ETA ) c(Zn) V (Zn)M(Al)w(A