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:
0.108
Explanation:
You use the formula Q =mco
M is mass
Q is th heat release
C is specific heat
O is the temperature change
Substitute what you have and what you get in the information in the equation then you will get your answers
Answer:
It's B !
Explanation:
Formulas. The molecular formula for glucose is C6H12O6. This means that there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms bonded together to make one molecule of glucose.
Hope this helps!!
Answer:
The mixture of cryolite and aluminum oxide has a lower melting point than pure aluminum oxide. This means a lower amount of energy is required to establish effective conditions for electrolysis and thus makes it more cost effective.
Explanation:
