Answer:
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Explanation:
2, 8,6 because it has to be in a configuration of 2,8,8
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Answer:
At 430.34 K the reaction will be at equilibrium, at T > 430.34 the
reaction will be spontaneous, and at T < 430.4K the reaction will not
occur spontaneously.
Explanation:
1) Variables:
G = Gibbs energy
H = enthalpy
S = entropy
2) Formula (definition)
G = H + TS
=> ΔG = ΔH - TΔS
3) conditions
ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction
4) Assuming the data given correspond to ΔH and ΔS
ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K
=> T = [ΔH - ΔG] / ΔS
ΔG = 0 => T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K
This is, at 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.
Answer:
6.93
Explanation:
correct me if verry wrong
