Answer:
C3H6O2
Explanation:
To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.
H = 4.10/1.36 = 3
O = 1.36/1.36 = 1
C = 2.05/1.36 = 1.5
We then multiply through by 2 to yield the compound with the empirical formula C3H6O2
Answer:
1. Gases can be easily liquefied into very small volumes and stored in liquid form Eg in LPGA cylinders and used in homes.
2. Balloons can be easily filled with air.
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
The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.
The volume of water is 
.
Since, 1 gal= 3785.41 mL
Thus, 
Density of water is 1 g/mL thus, mass of water will be 
.
Since, 1 grams of chlorine →
grams of water.
1 g of water →
g of chlorine and,
of water →86.6 g of chlorine
Since, the solution is 9% chlorine by mass, the volume of solution will be:
Thus, volume of chlorine solution is 9.62\times 10^{2} mL.
