How is iodine-131 the same as iodine-126
1 answer:
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Answer : Given data ;
Δ G° = 212 KJ/molTemperature is = 25+273 = 298 KAnd gas constant R = 0.008314 KJ/mol
The reaction is NiO(s) ⇌ Ni(s) +
We can find out the pressure on oxygen by using the equation of gibb's free energy with remainder quotient which is, ΔG = ΔG° +RT lnQ
when at equilibrium Q = K, here K is equilibrim constant, and ΔG becomes 0;
so we get, 0 = ΔG° + RT ln K
on rearranging we get, ln K = ΔG° / (RT)
ln K = 212 / (0.00831 X 298) = 85.6
Here, now K = = 1.51 X
When we get K = 1.51 X
But, K =
So, (1.51 X ) 
= 3.87 X 
Hence, the pressure of oxygen will be = 3.87 X Pa
Now, Pa to atm conversion will be 3.756 X
atm
Δ G° = 212 KJ/molTemperature is = 25+273 = 298 KAnd gas constant R = 0.008314 KJ/mol
The reaction is NiO(s) ⇌ Ni(s) +
when at equilibrium Q = K, here K is equilibrim constant, and ΔG becomes 0;
so we get, 0 = ΔG° + RT ln K
on rearranging we get, ln K = ΔG° / (RT)
ln K = 212 / (0.00831 X 298) = 85.6
Here, now K = = 1.51 X
When we get K = 1.51 X
But, K =
Hence, the pressure of oxygen will be = 3.87 X
Now, Pa to atm conversion will be 3.756 X
<u>Answer:</u> The value of is
<u>Explanation:</u>
We are given:
Initial moles of ammonia = 0.0280 moles
Initial moles of oxygen gas = 0.0120 moles
Volume of the container = 1.00 L
Concentration of a substance is calculated by:
So, concentration of ammonia =
Concentration of oxygen gas =
The given chemical equation follows:
<u>Initial:</u> 0.0280 0.0120
<u>At eqllm:</u> 0.0280-4x 0.0120-3x 2x 6x
We are given:
Equilibrium concentration of nitrogen gas =
Evaluating the value of 'x', we get:
Now, equilibrium concentration of ammonia =
Equilibrium concentration of oxygen gas =
Equilibrium concentration of water =
The expression of for the above reaction follows:
Putting values in above expression, we get:
Hence, the value of is