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Sloan [31]
3 years ago
7

The pressure and temperature inside a bike tire is 10 atm and 10k respectively. What will the pressure become in the tire when t

he temperature is increased 20k
Chemistry
1 answer:
igomit [66]3 years ago
3 0

Answer:

P_2=20atm

Explanation:

Hello,

In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:

\frac{P_1}{T_1}= \frac{P_2}{T_2}

Thus, we solve for the final pressure P2 to obtain it as shown below:

P_2=\frac{P_1T_2}{T_1}=\frac{10atm*20K}{10K} \\\\P_2=20atm

Hence, we notice that the temperature doubles as well as the pressure.

Best regards.

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The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has
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This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

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The given chemical reaction is:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate law expression is:

Rate=k[NO]^2[O_2]

Now we have to determine the number of molecules of NO\text{ and }O_2

In mixture 1 : There are 5 NO and 4 O_2 molecules.

In mixture 2 : There are 7 NO and 2 O_2 molecules.

In mixture 3 : There are 3 NO and 5 O_2 molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

Rate=k[NO]^2[O_2]

Rate=k(5)^2\times (4)

Rate=k(100)

The rate law expression for mixture 2 is:

Rate=k[NO]^2[O_2]

Rate=k(7)^2\times (2)

Rate=k(98)

The rate law expression for mixture 3 is:

Rate=k[NO]^2[O_2]

Rate=k(3)^2\times (5)

Rate=k(45)

Hence, the mixture 1 has the fastest initial rate.

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