MULTIPLE CHOICE

Chemistry

1 answer:

Flauer [41]3 years ago

6 0

Answer:

its very simple ans we have 2 just multiply256

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What is the mass in grams of 1 liter of nitrogen gas?

AlladinOne [14]

Assuming the volume of the gas is measured at standard temperature and pressure, Then one mole of the Gas would occupy 22.4 liters.

Therefore, 1 liter is 1/224 moles

one mole of nitrogen 14 is 14

Therefore 1 liter of the nitrogen weighs 1/224×14

0.0625 grams

7 0

3 years ago

3. Which one is part of the kinetic molecular theory? Yes = part of KMT, No = not part

lidiya [134]

Explanation:

a. Average kinetic energy is directly proportional to absolute Kelvin temperature of a gas.

Yes

b. There are no attractive forces and repulsive forces between gas molecules.

Yes

C. Atoms are neither created nor destroyed by ordinary chemical reactions.

d. The volume occupied by all of the gas molecules in a container is negligible compared to the volume of the container

Yes

The kinetic molecular theory is one of such theories used to explain the forces between molecules and the energy they posses.

According to the theory;

The temperature of gas is proportional to the average kinetic energy.
Molecules are independent of one another and the force of attraction and repulsion between them is negligible.
volume occupied by gases is negligible compared to the volume of the container.

Law of conservation of matter states that "atoms are neither created nor destroyed by ordinary chemical reactions".

learn more:

Kinetic molecular theory brainly.com/question/12362857

#learnwithBrainly

6 0

3 years ago

the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh

Harman [31]

Answer:

$\boxed{\text{10 L}}$

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

$\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

$\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}$

Calculation:

$\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}$