Before the bullet is fired the momentum is Zero because nothing is moving but once the bullet is shot the momentum increases because of the movement of the bullet moving forward.

8 0

3 years ago

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat

Valentin [98]

Answer:

(a)

$0.0342M$

(b)

$t_{1/2}=17.36s\\t_{1/2}=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\$

$[NOBr]=\frac{1}{29.2/M}=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_{1/2}=\frac{1}{k[NOBr]_0}$

Therefore, for the given initial concentrations one obtains:

$t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s$

Best regards.

8 0

3 years ago

Read 2 more answers

In the United States, barometric pressures a re generally reported in inches of mercury (in. Hg). On a beautiful summer day in C

Svetllana [295]

Answer:

773.43 torr

Explanation:

From the question given above, the following data were obtained:

Pressure (in inHg) = 30.45 inHg

Pressure (in torr) =?

We can convert 30.45 inHg to torr by doing the following:

1 inHg = 25.4 torr

Therefore,

30.45 inHg = 30.45 inHg × 25.4 torr / 1 inHg

30.45 inhg = 773.43 torr

Thus, 30.45 inhg is equivalent to 773.43 torr

7 0

3 years ago