C3H8 (g) +5O2 (g) =3CO2 (g) +4H2O (l); △H 1=-22 15.0？ KJ/mol, water is a * * * valence compound, and the electronic type is:
So the answer is: c3h8 (g)+5o2 (g) = 3co2 (g)+4h2o (l); △h 1 =-22 15.0 kJ/mol；
② 1mol dimethyl ether is completely combusted to generate CO2, and liquid water gives off heat of 1455kJ. If the mixed gas of propane and dimethyl ether at 1mol is completely combusted to generate CO2, and the liquid water releases heat at 1645kJ, let the amount of dimethyl ether in the mixed gas at 1mol be X and the amount of propane be1-X. △ h1=-22/kloc. The heat release of propane combustion is (1-0). According to the conditions:1645kj-1455xkj = (1-x) 2215kj, x=0.75, then the amount of mixed propane is 0.25mol.
(2)① The reaction A+B → C (△ H < 0) is carried out in two steps. 1a+b → x (△ h > 0) 2x→ c (△ h < 0) shows that A+B → C (△ H < 0) is an exothermic reaction, and the sum of energies of a and b is greater than c, which is determined by 1a+. The energy of A+B is greater than that of c; The energy of x is greater than that of c, which is consistent with d.
So the answer is: d:
(2) Calculation according to Gus's law;
①C2H5OH (g) +3O2 (g) =2CO2 (g) +3H2O (g); △H 1=-Q3 kJ/mol，
②C2 H5 oh(g)= C2 H5 oh(l)； △H2=-Q2 kJ/mol,
③H2O(g)= H2O(l)； △H3=-Q 1 kj/mol.
According to Gaith's law: ①-②+3x③, we get: C2H5OH (L)+3O2 (G) = 2CO2 (G)+3H2O (L); △h 1 =-(Q3-Q2+3q 1)kJ/mol，
If the amount of 23g liquid anhydrous ethanol is 0.5mol, and it returns to room temperature after complete combustion, the heat released in the whole process is (0.5Q3-0.5Q2+1.5q1) kj;
So the answer is: 0.5Q3-0.5Q2+1.5q1;
(3) Design an experiment to calculate the △H of C(s)+ 1/2O2(g)=CO(g) by using Gaith's law. It is necessary to know the combustion heat of carbon and carbon monoxide.
So, the answer is: the heat of combustion of carbon and carbon monoxide;