Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
It would be MnSO4
The (II) lets you know it’s the form with a 2+ charge and Sulfate has a 2- charge
These will cancel out making it plain MnSO4
If it was manganese (iii) sulfide the answer would be Mn2(SO4)3
