Question 7 (I’m a bit rusty on this so may be wrong):
In order for everything to cancel out, we firstly need to make sure all the products are present when the reactions are added, which means we need to triple reaction 2 in order for it to match 3H2O. Then, we need the reactants to match the equation which means we need to reverse the first one, since B2H6 is on the LHS in the intended equation. So, adding the negative of the first enthalpy, triple the second enthalpy and then the third enthalpy, we get:
-(35.75) + 3(-286) + (-1274) = -2167 kJmol
Someone else can help with the rest I’m tired