elielson1497

25.11.2021 •

Chemistry

the absence of c5 means that the molecule is not present to be activated. what happens in normal serum when c5 is ""activated""?

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Ответ:

smesarahmitchell

22.08.2020

Chemistry

$\boxed{{\text{103}}{\text{.94 mL}}}$ of gas exerts a pressure of 457 mm Hg.

Further Explanation:

Ideal gas:

An ideal gas contains a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is just a theoretical concept, and practically no such gas exists. But gases tend to behave almost ideally at a higher temperature and lower pressure.

Ideal gas law is considered as the equation of state for any hypothetical gas. The expression for the ideal gas equation of gas is as follows:

${\text{PV}}={\text{nRT}}$ …… (1)

Here,

P is the pressure of the gas.

V is the volume of gas.

n denotes the number of moles of gas.

R is the gas constant.

T is the temperature of gas.

Boyle’s law:

It is an experimental gas law that describes the relationship between pressure and volume of the gas. According to Boyle's law, the volume of the gas is inversely proportional to the pressure of the system, provided that the temperature and the number of moles of gas remain constant.

If the temperature and number of moles of gas are constant then the equation (1) will become as follows:

${\text{PV}}={\text{constant}}$ …… (2)

Or it can also be expressed as follows:

${{\text{P}}_1}{{\text{V}}_1}={{\text{P}}_2}{{\text{V}}_2}$ …… (3)

Here,

${{\text{P}}_1}$ is the initial pressure.

${{\text{P}}_2}$ is the final pressure.

${{\text{V}}_1}$ is the initial volume.

${{\text{V}}_2}$ is the final volume.

Rearrange the equation (3) for ${{\text{P}}_2}$ , and we get,

${{\text{P}}_2}=\frac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{V}}_2}}}$ …… (4)

The conversion factor to convert ${\text{atm}}$ into ${\text{mm Hg}}$ is as follows:

${\text{1 atm}}={\text{760 mm Hg}}$

The value of ${{\text{P}}_1}$ is 2.50 atm.

The value of ${{\text{V}}_1}$ is 25 mL.

The value of ${{\text{P}}_2}$ is 457 mm Hg.

Substitute these values in equation (4).

$\begin{aligned}{{\text{V}}_{\text{2}}}&=\frac{{\left( {{\text{2}}{\text{.50 atm}}}\right)\left({\frac{{760\;{\text{mmHg}}}}{{{\text{1}}\;{\text{atm}}}}}\right)\left({{\text{25 mL}}}\right)}}{{\left( {457\;{\text{mmHg}}}\right)}}\\&= {\text{103}}{\text{.938731 mL}}\\&\simeq {\text{103}}{\text{.94 mL}}\\\end{aligned}$

The final volume of a gas is 103.94 mL.

Learn more:

1. Determine the final pressure of the ideal gas:

2. Chemical bonds in NaCl:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: ideal gas, pressure, volume, absolute temperature, equation of state, hypothetical, universal gas constant, moles of gas, initial, final.

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