化學試題可以告訴我計算過程嗎？

Question

1.A 2.839-g sample of C2H4O was burned in a bomb calorimeter with a total heat capacity of 16.77 kJ/°C. The temperature of the calorimeter increases from 22.62°C to 26.87°C. What is the heat of combustion per mole of C2H4O?
answer--1.10 x 10三次方 kJ/mol 計算過程是？
2.Lactic acid, HC3H5O3, has one acidic hydrogen. A 0.10 M solution of lactic acid has a pH of 2.44. Calculate Ka.
answer--1.4 x 10負4次方
3.The thermal decomposition of N2O5(g) to form NO2(g) and O2(g) is a first-order reaction. The rate constant for the reaction is 5.1 x 10-4 s-1 at 318 K. What is the half-life of this process?
4.Complete and balance the following equation. (All stoichiometric coefficients must be integers.)
MnO4-(aq) Cl-(aq) Mn2 (aq) Cl2(g) (acidic solution)

How many hydrogen ions are needed and on which side of the equation must they appear?
5.What is the final pH if 0.020 mol HCl is added to 0.500 L of a 0.28 M NH3 and 0.22 M NH4Cl buffer solution (Kb (NH3) = 1.8 x 10-5)?

Answer 1

16.77x(26.87-22.62)/2.839x44 = 1.1x103 kJ/molpH = 2.44 => [H+] = 10-2.44 M   Ka = [10-2.44]2/(0.1-10-2.44) = 1.4x10-4 2N2O5 --> 4NO2 + O2  R = k[O2]2MnO4-(aq) + 10Cl-(aq) + 16H+ --> 2Mn2+(aq) + 5Cl2(g) + 8H2ONH3 － 0.5x0.28 = 0.14 molNH4+ － 0.5x0.22 = 0.11 mol加入 0.020 mol HCl ，則 NH3 和 NH4+ 的莫耳數改變為－NH3 － 0.14-0.02 = 0.12 mol  [NH3] = 0.12/0.5 = 0.24 MNH4+ － 0.11+0.02 = 0.13 mol  [NH4+] = 0.13/0.5 = 0.26 M

2006-01-05 20:43:08 補充：
第五題：
NH3 － 0.5x0.28 = 0.14 mol
NH4+ － 0.5x0.22 = 0.11 mol
加入 0.020 mol HCl ，則 NH3 和 NH4+ 的莫耳數改變為－
NH3 － 0.14-0.02 = 0.12 mol [NH3] = 0.12/0.5 = 0.24 M
NH4+ － 0.11+0.02 = 0.13 mol [NH4+] = 0.13/0.5 = 0.26 M

2006-01-05 20:43:55 補充：
而解離方程式為　NH3 + H2O -- NH4+ + OH-
Kb = [NH4+][OH-]/[NH3] = 0.26x[OH-]/0.24 = 1.8x10-5
[OH-] = 1.66x10-5
pOH = 4.78
pH = 9.22