Volume can be measured in cubic centimetres (which is the small 3 next to the unit and is higher)
Answer:
-252.5 kJ/mol = ΔH H2O(g)
Explanation:
ΔH Fe2O3 = -825.5kJ/mol
ΔH H2 = 0kJ/mol
ΔH Fe = 0kJ/mol
Based on Hess's law, ΔH of a reaction is the sum of ΔH of products - ΔH of reactants. For the reaction:
Fe2O3(s) + 3 H2(g) →2Fe(s) + 3 H2O(g)
ΔHr = 67.9kJ/mol = 3*ΔH H2O + 2*ΔHFe - (ΔH Fe2O3 + 3*Δ H2)
67.9kJ/mol = 3*ΔH H2O + 2*0kJ/mol - (ΔH -825.5kJ/mol + 3*Δ H2)
67.9 = 3*ΔH H2O(g) + 825.5kJ/mol
-757.6kJ/mol = 3*ΔH H2O(g)
<h3>-252.5 kJ/mol = ΔH H2O(g)</h3>
Answer:
Tc
Explanation:
You just have to follow the rows with the exponents. Just remember that when we get to d, the number in the front is a period lower. Hope this helps!
Answer:
Molecular formula: C₂H₄O₂
Empirical formula: CH₂O
Explanation:
40 % C, 6.72 % H and 53.29 % O states the centesimal composition of the compound. These data means that in 100 g of compound we have x grams of a determined element.
We divide the mass by the molar mass of each:
40 g / 12 g/mol = 3.33 moles of C
6.72 g / 1 g/mol = 6.72 moles of H
53.29 g / 16 g/mol = 3.33 moles of O
We can determine rules of three to get, the molecular formula.
In 100 g of compound we have 3.33 moles of C, 6.72 moles of H and 3.33 moles of O; therefore in 60 g (1 mol) we must have
- (60 . 3.33) / 100 = 2 moles of C
- (60 . 6.72) / 100 = 4 moles of H
- (60 . 3.33) / 100 = 2 moles of O
Molecular formula is C₂H₄O₂
Empirical formula has the lowest suscripts; we divide by two, so the empirical formula is CH₂O
Answer:
reactive. hope this helps
