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>
when Kc = [the concentration of products]^(no.of its mol in the balanced equation) / [the concentration of the reactants]^(no.of mol in the balanced equation)
∴ The equilibrium constant expression for this reaction is:
Kc= [H+]^2 * [So4-2] / [H2SO4]
W = F * s
W = 100 * 10
W = 1000 J
In short, Your Answer would be 1000 Joules
Hope this helps!
Answer:
b melting
Explanation:
The phase change the substance is undergoing is from solid to liquid and this is called melting.
- To understand this process, we need to know the the substance changing.
- A substance that contains particles that are not easily compressed and packed is a solid.
- Solids have fixed shapes and volumes.
- Liquids can easily flow although they cannot also be easily compressed.
The phase change from solid to liquid is facilitated by a melting process.
