01緒論第1章分析化學概論seen

1、1入 0.5000mol/LH2SO4 c1V1c2V2 c1入 0.5000mol/LH2SO4 c1V1c2V2 c(V1 V0.0982mol / L0.480L 0.5000mol / LV2 0.1000mol / L(0.480L VV2 甲酸氫鉀(KHC8H4O4)？n:8 4 m1n1M 0.2mol/L0.025L204.22g/mol m2 n2M 0.2mol/L0.030L204.22g/mol NaOH HCO2H2 2 4 m nM 1cV1210.2mol/L0.025L126.07g/mol 0.3g m n M 1cV 2210.2mol/L0.030L126

2、.07g/mol 0.4g 6含S 有機試0.471gS 氧化SO2，用預中和過的 H2SO4， 將28.2mLSS6含S 有機試0.471gS 氧化SO2，用預中和過的 H2SO4， 將28.2mLSSSO2 H2SO4 w nM 10.108mol/L0.0282L32.066g/ 100%80.2500g溶解，煮沸除CO2，用0.2450mol/LNaOH 溶液反滴定過量酸，消CaCO3 CaCO3 NaOH1(cV cV2w nM 100%1 (0.2600mol / L0.025L0.2450mol / L0.0065L)100.09g / 100% 使 Fe3+還原 SbSb2S3

3、 2Sb3SO2 6Fe2+ 5cV 50.0200mol/L0.03180L 662nSb S 20.00053mol 2 0.00053mol339.68g/mol100%w 0.00106mol121.76g/ 5cV 50.0200mol/L0.03180L 662nSb S 20.00053mol 2 0.00053mol339.68g/mol100%w 0.00106mol121.76g/mol 100%w12. 與36.42mLKMnO4 KMnO4 5As2O3 10AsO33- 4MnO4- 4 m 40.2112 0.02345(mol/c36.42作為還原劑?？煞磻?H

4、C O +2MnO- +6H+=10CO +2Mn2+8H 2 2 422 m2 2 4 H C 5.55310 nMQ2nHCO 211.10610 2 2 11.106n 0.111(L) 2QH C 2 2 45 25 25.553103 2.221103(mol) n 0.0222(L)Vc含K2Cr2O75.442gL-1（M=231.54gmol-1）的滴定度 25 25.553103 2.221103(mol) n 0.0222(L)Vc含K2Cr2O75.442gL-1（M=231.54gmol-1）的滴定度（mg/mL）nK Cr 2 2 7 294.18 0.01850(m

5、ol/cV1 M2 0.01850231.542 8.567(mg /TFe O /K Cr K Cr Fe 3 4 2 2 2 2 3 FeSO47H2O（M=278.04gmol-1）的含量18. 99.50100.5%為一級（G.R）；99.00%100.5%為二級為三級（C.P）?，F(xiàn) FeSO47H2O 解： 5Fe2MnO4 5 35.70 0.02034 3.631103FeSO74MmnFeSO 7H FeSO 7H FeSO 7H FeSO4 7mm3.631103278.04 20. CN-可Ni(CN4 Ni2+EDTA標準溶液滴定，Ni(CN4 CN-解Ni2+39.3nn CN-解Ni2+39.3nn 0.01703(m