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Sunny_sXe [5.5K]
3 years ago
12

what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s

?
Physics
1 answer:
yaroslaw [1]3 years ago
6 0

Answer : The value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

Explanation :

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant = ?

t = time = 17s

[A_t] = final concentration = 0.0981 M

[A_o] = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}

k=0.51M^{-1}s^{-1}

Therefore, the value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

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Answer:

(d) 0.07 m/s

Explanation:

Given Data

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From law of conservation of momentum

m_{1}v_{1}=m_{2}v_{2}\\  v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s

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3 years ago
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