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Answer:
4054 kcal of heat is released during complete combustion of 354 g of octane.
Explanation:
Heat of combustion of 1 mol of octane is
kcal
Molar mass of octane = 114.23 g/mol
We know, no. of moles = (mass)/(molar mass)
So,
kcal of heat is released during complete combustion of 114.23 g of octane.
So, amount of heat is released during complete combustion of 354 g of octane =
kcal = 4054 kcal
Hence 4054 kcal of heat is released during complete combustion of 354 g of octane.
Answer:
-490.7 K
Explanation:
Given:
[Ni^2+]= 0.4 M
[Pb^2+]=0.002 M
∆V= -0.012 V
VNi= -0.250V
VPb= -0.126V
F= 96500 C
R= 8.314 JK-1 mol-1
n= 2
From
T= -nF/R [∆V-(VNi-VPb)/ln [Pb2+]/[Ni2+]]
T= 2(96500)/8.314[ (-0.012) -(-0.250) - (-0.126))/ln[0.002]/[0.4]
T= 23213.856(0.112/(-5.298))
T= -490.7 K
A rapid release of stored up energy
Answer:
Ca
2+
<K + <Ar<Cl − <S 2−
Explanation:
Ar,K +
,Cl −
,S 2−
,Ca 2+
have the same number of electrons. Their radii would be different because of their different nuclear charges. The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size. Hence the correct order will be Ca
2+ <K + <Ar<Cl − <S 2−