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tatiyna
4 years ago
5

A force has to have which two Factors

Chemistry
2 answers:
HACTEHA [7]4 years ago
7 0
Hiya!
Its B Magnitude and Direction!

Hope This Helps!
(I did the work) :)
monitta4 years ago
4 0
I think that the answer is a size and speed
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At 450 K the rate constant is 15.4 atm-1s-1. How much time (in s) is needed for NOCl originally at a partial pressure of 56 torr
Gwar [14]

This is an incomplete question, here is a complete question.

Consider the second-order decomposition of nitroysl chloride:

2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)

At 450 K the rate constant is 15.4 atm⁻¹s⁻¹. How much time (in s) is needed for NOCl originally at a partial pressure of 53 torr to decay to 10.6 torr?

Answer : The time needed for NOCl is, 2.52 seconds.

Explanation :  Given,

Rate constant = 15.4atm^{-1}s^{-1}

Initial partial pressure of NOCl = 56 torr = 0.0737 atm

final partial pressure of NOCl = 14.5 torr = 0.0191 atm

The expression used for second order kinetics is:

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

where,

k = rate constant

t = time

[A_t] = concentration at time 't'

[A_o] = initial concentration

As we know that,

PV=nRT\text{ or }PV=CRT

Thus, the expression of second order kinetics will be:

kt=RT\times \frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}

\frac{k}{RT}t=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}

As, k'=\frac{k}{RT}

So, k't=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}         ............(1)

Now put all the given values in the above expression 1, we get:

(15.4atm^{-1}s^{-1})\times t=\frac{1}{0.0191atm}-\frac{1}{0.0737atm}

t=2.52s

Therefore, the time needed for NOCl is, 2.52 seconds.

5 0
3 years ago
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