(2) The valence of C in urea is +4, and only N changes valence in the reaction. In 4 moles of CO (NH2) 2, -3 valence N loses 24 moles of electrons, and in 6 moles of NO 2, +4 valence N gains 24 moles of electrons, resulting in 7 moles? N2*** transfers 24mol of electrons, so the amount of substance 12g = 12g60g/mol=0.2mol, so when 12g of urea is consumed, * * transfers 1.2? Mol electron, so the answer is: 1.2? mol
(3)6NO (g) +4NH3 (g)? 5N2 (g) +6H2O (g) △ h1=-1627.2kjmol-1①
4NO (g) +4NH3 (g) +O2 (g)? 4N2 (g) +6H2O (g) △ H2 =-18070.0 kj mol-1②
6NO (g) +8NH3 (g)? 7N2 (g)+12H2O (g) △H3=-2659.9kJmol- 1③
Equation ②-① gives N2 (g)+oxygen (g)? 2NO(g)△H =△H2-△H 1 =- 18070.0 kj mol- 1-(- 1627.2 kj mol- 1)=- 179.8 kj/mol，
So the answer is:-179.8;
(4) According to the image, the temperature rises and the balance moves in the opposite direction, so △ h < 0; When T 1, α=0.5 corresponding to point A. According to the expression of equilibrium constant, K = 1 can be obtained. Because the temperature at point B and point A is the same, K is constant, and the volume can be calculated as 2L, so the answer is:
(5) When NH4+ in (NH4) 2SO4 solution is hydrolyzed into acidity, the order of ion concentration is C (NH4+) > C (SO4 2-) > C (H+) > C (OH-);
0.05mol？ L- 1(NH4)2SO4 solution with pH=a, then c(OH-)= 10a- 14, Kb=c(NH4+)? C (oh? )c(NH3？ H2O) 1.7× 10-5 mol. L- 1, so c(NH4+)c(NH3? H2O)= 1.7× 10？ 5 1× 10a？ 14= 1.7× 109-a，
So the answer is: C (NH4+) > C (SO4 2-) > C (H+) > C (OH-); 1.7× 109-a