How many electrons does each gold atom gain? How many electrons does each iodine atom lose? What is the total number of electron
2 answers:
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Answer: The half-life of a first-order reaction is,
Explanation:
All the radioactive reactions follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
where,
k = rate constant = ?
t = time taken = 440 s
= initial amount of the reactant = 0.50 M
[A] = left amount = 0.20 M
Putting values in above equation, we get:
The equation used to calculate half life for first order kinetics:
Putting values in this equation, we get:
Therefore, the half-life of a first-order reaction is,
Answer : The concentration of at equilibrium is 0.9332 M
Solution : Given,
Concentration of and
at equilibrium = 0.200 M
Concentration of and
at equilibrium = 0.500 M
First we have to calculate the value of equilibrium constant.
The given equilibrium reaction is,
The expression of will be,
Now we have to calculate the final concentration of NO.
The given equilibrium reaction is,
Initially 0.200 0.200 0
.800
At equilibrium (0.200-x) (0.200-x) (0.800+2x)
The expression of will be,
By solving the term x, we get
From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.
And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).
Thus, the concentration of at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M
Therefore, the concentration of at equilibrium is 0.9332 M