How much heat is released when 1.4 mol of hydrogen fluoride are produced? H2(g) +F2(g) → 2HF(g) +536kJ

(Please show work, Thanks!)

+375.2 KJ.

Explanation:

The balanced equation for the reaction is given below:

H₂ + F₂ —> 2HF ΔH = +536 KJ

From the balanced equation above,

2 moles of HF required +536 KJ .

Finally, we shall determine the heat required to produce 1.4 mol of hydrogen fluoride, HF. This is illustrated below:

From the balanced equation above,

2 moles of HF required +536 KJ .

Therefore, 1.4 moles of HF will require = (1.4 × 536)/2 = +375.2 KJ

Thus, +375.2 KJ of heat energy is required.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o