Answer:Yes they are in the same mineral group
Explanation:zinc is the central elements there. The rest of the elements are present as impurities due to where it was found. Like carbon is can be found in the soil, silicon with oxygen is basically sand, hydrogen is in the atmosphere and also in water and soil too. So apart from zinc, the rest are normal day to day elements.
Answer:
Zero
Explanation:
FrBr is an ionic compound
.
Fr is in Group 1. Br is in Group 17.
The charges on the ions are +1 and -1, respectively.
The compound consists of Fr⁺Br⁻ ions.
However, there are equal numbers of + and - charges, so
The overall charge of the compound is zero.
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
Answer:
a) 0.210 j
/k
b) 0.032 j/k
Explanation:
Find the attachment for solution
