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:
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Answer:
mass Na2SO4 = 14.3816 g
Explanation:
sln Na2SO4:
∴ V = 450 mL
∴ <em>C </em>=<em> </em>0.2250 mol/L
∴ Mw ≡ 142.04 g/mol..... from literature
⇒ mol Na2SO4 = (0.2250 mol/L)(0.450 L) = 0.10125 mol
⇒ mass Na2SO4 = (0.10125 mol)(142.04 g/mol) = 14.3816 g