An object's shadow was observed and measured at different times of the day. The data was recorded in the chart below. How does the data provide evidence that the sun is changing position in the sky throughout the day?

The reaction enthalpy for the oxidation of 1 mole of glucose is -1,055 kJ.

Explanation:

The standard enthalpy of a reaction can be calculated from the standard enthalpies of formation of each substance, using the following expression:

ΔH°r = ∑n(p).ΔH°f(p) - ∑n(p).ΔH°f(p)

where,

ΔH°r is the standard enthalpy of the reaction

n(i) is the number of  moles of reactants and products in the balanced equation

ΔH°f(i) is the standard enthalpy of formation of reactants and products, which are tabulated.

Considering the balanced equation:

2 O₂(g) + C₆H₁₂O₆(s) → 2 CH₃COOH(l) +2 CO₂(g) +2 H₂O(l)

The standard enthalpy of reaction is:

ΔH°r = [2 x ΔH°f(CH₃COOH(l)) + 2 x ΔH°f(CO₂(g)) + 2 x ΔH°f(H₂O(l))] - [2 x ΔH°f(O₂(g)) + 1 x ΔH°f(C₆H₁₂O₆(s))]

ΔH°r = [2 mol x (-483.5 kJ/mol) + 2 mol x (-393.5 kJ/mol) + 2 mol x (-285.8 kJ/mol)] - [2 mol x 0 kJ/mol + 1mol x (-1,271 kJ/mol)]

ΔH°r = -1,055 kJ

In the balanced equation there is 1 mole of glucose. So, the enthalpy of reaction for such amount is -1,055 kJ (exothermic).