Someone plz help me here are the photos, I need the last one

(6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol) - (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)

Explanation:

Question; From the given options, the chemical reaction in the question is presented as follows;

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), given that we have;

for C₆H₁₂O₆ = -1,273.02 kJ/mol

for O₂(g) = 0 kJ/mol

for CO₂(g) = -393.5 kJ/mol

for H₂O(l) = -285.83 kJ/mol

The heat or enthalpy of a reaction,  is given as follows;

Therefore, the equation which should be used to calculate , is given as follows;

= (6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol)

= (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)

Therefore;

= (6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol) - (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.