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:
Farther, longer, longer
Explanation:
The further away a planet is from the sun, the larger the orbit. This makes the time for the planet to make a revolution around the sun longer.
Answer:
Explanation:
use the equation
moles = mass/mr
=19.9/79.5
=0.250moles of CuO
then do the same for
H = 2.02/1
=2.02
so CuO is the limiting reagent because there is less amount of it.
Hope this helps :)
(0.48 gram) x (1mole / 4.0 gram) = 0.48/4.0 = 0.12 mole
