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3 years ago

How does one's lifestyle contribute to the increase in atmospheric CO2 concentration?

Chemistry

1 answer:

Black_prince [1.1K]3 years ago

4 0

It contributes by breathing. CO2 is carbon dioxide we breath it out and plants convert it into oxygen.

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12.70 L of a gas has a pressure of 0.63 atm. What is the volume of the gas at 105kPa?

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Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=0.63atm\\V_1=12.70L\\P_2=105kPa=1.04atm(1kPa=0.009atm)\\V_2=?$

Putting values in above equation, we get:

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3 years ago

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You used a calorimeter in the heat transfer lab. Explain how the calorimeter works, and how to calculate the heat given off or a

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3 years ago

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha

sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

Given:

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

To determine:

The mean free path, λ

Calculation:

The mean free path is related to the collision cross section by the following equation:

$\lambda =\frac{1}{n\sigma }------(1)$

where n = number density

$n = \frac{P}{kT}-----(2)$

Substituting for P, k and T in equation (2) gives:

$n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\ m^{3}$

Next, substituting for n and σ in equation (1) gives:

$\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m$

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3 years ago

How does one's lifestyle contribute to the increase in atmospheric CO2 concentration? ​

How does one's lifestyle contribute to the increase in atmospheric CO2 concentration?