Me this 2 . i’m not good at this sorry

Nitric oxide (no) can be formed from nitrogen, hydrogen and oxygen in two steps. in the first step, nitrogen and hydrogen react to form ammonia: n2(g) + 2 h_2(g) rightarrow 2 nh_3 (g) delta h = -92. kj in the second step, ammonia and oxygen react to form nitric oxide and water: 4 nh_3(g) + 5 o_2(g) rightarrow 4no(g) + 6 h_2o(g) delta h = -905. kj calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. round your answer to the nearest .

If 10.0 ml of the solution on the right are withdrawn from the 100 ml beaker and diluted again in a similar manner, what is the new concentration? m nacl

For the reaction, 4 a(g) + 3 b(g) => 2 c(g), the following data were obtained at constant temperature. experiment initial[a],mol/l initial [b],mol/l initial rate,m/min 1 0.200 0.150 5.00 2 0.400 0.150 10.0 3 0.200 0.300 10.0 4 0.400 0.300 20.0 which of the following is the correct rate law for the reaction? 1. rate = k[a]2[b]2 2. rate = k[a][b] 3. rate = k[a]2[b] 4. rate = k[a][b]2