The value of δh° for the reactiohe value of δh° for the reaction below is -790 kj. the enthalpy change accompanying the reaction of 0.95 g of n below is -790 kj. the enthalpy change accompanying the reaction of 0.95 g of s is kj.

Explanation:

You know how much S you have, so convert that to moles by dividing it with its atomic mass of 32g/mol

(0.95g S)/(32 mol S/g S)  =0.0297mole

then, you know for each 2 mol of S you put into the reaction, you get -790kJ so:

(0.0297mole)(-790kJ/2 mol S)= -11.7kJ

i don't have a calculator handy so you can just do that on your calculator.

so what is the problem with your lab report and whats the subject of the lab report? i will you