The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
hope it helps.
<h3>stay safe healthy and happy.<u>.</u><u>.</u></h3>
D. One chimp cleaning and grooming the hair of another chimp
The mass of melted gold to release the energy would be 3, 688. 8 Kg
<h3>How to determine the mass</h3>
The formula for quantity of energy is given thus;
Q = n × HF
Where n represents number of moles
HF represents heat of fusion
To find the number of moles, we have
235.0 = n × 12.550
number of moles = 
= 18. 725 moles
Note that molar mass of Gold is 197g/ mol
Let's note that;
Number of moles = mass/ molar mass
Mass = number of moles × molar mass
Mass = 18. 725 × 197
Mass = 3, 688. 8 Kg
Thus, the mass of melted gold to release the energy would be 3, 688. 8 Kg
Learn more about molar heat of fusion here:
brainly.com/question/15634085
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