Answer: 0.0043mole
Explanation:Please see attachment for explanation
<h3>The average atomic mass of Iodine : 126.86 amu</h3><h3>Further explanation</h3>
Given
80% 127I, 17% 126I, and 3% 128I.
Required
The average atomic mass
Solution
The elements in nature have several types of isotopes
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.% + ... mass isotope n.%
Atomic mass of Iodine = 0.8 x 127 + 0.17 x 126 + 0.03 x 128
Atomic mass of Iodine = 101.6 + 21.42 + 3.84
Atomic mass of Iodine = 126.86 amu
Heat is given by multiplying the specific heat capacity of a substance by mass and the change in temperature. The heat capacity of water is Approximately 4184 J/K/C.
Therefore, heat = mc0 mass in kg
= (422/1000) × 4184 × (100-23.5)
= 135072.072 J
Latent heat of vaporization is 2260 kJ/kg
Thus the heat will be 0.422 × 2260000 = 953720 J
Heat to raise steam from 100 to 150
2000 × 0.422 ×50 = 42200 J
Thus the heat required is (135072.072 + 953720 + 42200) = 1330992.07 Joules or 1330 kilo joules
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