The most abundant element in the Sun and in the stars are hydrogen and helium. Like most of the stars, there is a spontaneous radioactive reaction happening in the Sun. Hydrogen is transformed into Helium. As long as the stars are young, the most abundant element is hydrogen.
To make it easier, assume that we have a total of 100 g of a compound. Hence, we have 58.80g of xenon, 7.166g of oxygen, and 34.04g of fluorine.
Know we will convert each of these masses to moles by using the atomic masses:
58.8/131.3 = 0.45 mole of Xe
7.166/16 = 0.45 mole of O
34.04/19 = 1.79 mole of F
Now, we will divide all the mole numbers by the smallest among them and get the number of atoms in the compound:
Xe = 0.45/0.45 = 1
O = 045/0.45 = 1
F = 1.79/0.45 = 3.98 = 4
So, the empirical formula of the compound XeOF₄
Answer:The standard enthalpy of combustion per gram of liquid methanol is 45.40kJ/g
Explanation:
Standard enthalpy of combustion of methanol :
The standard enthalpy of combustion per gram of liquid methanol:
The standard enthalpy of combustion per gram of liquid methanol is 45.40kJ/g
Answer:
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
(Ans)
ΔHf° of CaC2 = -59.0 kJ/mol
Explanation:
CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ
ΔHrxn = −127.2kJ
ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);
ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn
Where
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol
ΔHf°(CaC2) = -59.0 kJ/mol