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3 years ago

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If He(g) has an average kinetic energy of 4790 J/mol under certain conditions, what is the root mean square speed of Cl2(g) mole

cules under the same conditions?

1 answer:

Angelina_Jolie [31]3 years ago

7 0

We need to know the root mean square speed of Cl₂ molecule.

The root mean square speed of Cl₂ molecule is:3.67 X 10⁴ cm/sec.

Average kinetic energy= $\frac{3}{2}$ RT, where R=Molar gas constant and T=absolute temperature in Kelvin.

Given, average kinetic energy= 4790 J/mol, R= 8.314 J/mol/K

So, T= $\frac{4790 X 2}{3 X 8.314}$

T=384.09 K

Root mean square velocity= $\sqrt{\frac{3RT}{M} }$

For, Cl₂ molecule, M= 71 g/mol, T=384.09 K (as condition is same as previous)

So, Root mean square velocity= $\sqrt{\frac{3 X 8.314 X 10^{7} X 384.09}{71} }$ =3.67 X 10⁴ cm/sec

Hence, Root mean square velocity= 3.67 X 10⁴ cm/sec.

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A sample of nitrogen gas is at a temperature of 50 c and a pressure of 2 atm. If the volume of the sample remains constant and t

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Answer:

The new temperature of the nitrogen gas is 516.8 K or 243.8 C.

Explanation:

Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

P1= 2 atm
T1= 50 C= 323 K (being 0 C= 273 K)
P2= 3.2 atm
T2= ?

Replacing:

$\frac{2 atm}{323 K} =\frac{3.2 atm}{T2}$

Solving:

$T2*\frac{2 atm}{323 K} =3.2 atm$

$T2=3.2 atm*\frac{323 K}{2 atm}$

T2= 516.8 K= 243.8 C

<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>

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Over long time scales the solubility-temperature feedback (below), can affect climate. An increase in atmospheric CO2 concentrat

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The fact that CO2 is released from oceans due to further rise in temperature is an example of a negative feedback.

A negative feedback is one in which the process that produces the feedback is interrupted. That is, the process is stopped as a result of the feedback received.

In this case, CO2 which leads to global warming dissolves in the ocean which serves a large sink for the gas. However, as the increase in ocean temperatures causes decrease in solubility of CO2, more CO2 is released leading to further temperature rise. This is an example of a negative feedback loop.

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27) Partial pressure of oxygen: 57.8 kPa

29) Final volume: 80 mL

30) Final volume: 8987 L

31) Due to property of water of being polar, ice floats on water

Explanation:

27)

In a mixture of gases, the total pressure of the mixture is the sum of the partial pressures:

$p_T = p_1 + p_2 + ... + p_N$

In this problem, the mixture contains 3 gases (helium, carbon dioxide and oxygen). We know that the total pressure is

$p_T=201.4 kPa$

We also know the partial pressures of helium and carbon dioxide:

$P_{He}=125.4 kPa\\P_{CO_2}=18.2 kPa$

The total pressure can be written as

$p_T=p_{He}+p_{CO_2}+p_{O_2}$

where $p_{O_2}$ is the partial pressure of oxygen. Therefore, we find

$p_{O_2}=p_T-p_{He}-p_{CO_2}=201.4-125.4-18.2=57.8 kPa$

29)

Assuming that the pressure of the gas is constant, we can apply Charle's law, which states that:

"For an ideal gas at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically,

$\frac{V}{T}=const.$

where

V is the volume of the gas

T is the Kelvin temperature

We can re-write it as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Here we have:

$V_1 = 42 mL$ (initial volume)

$T_1=-89^{\circ}C+273=184 K$ is the initial temperature

$T_2=77^{\circ}C+273=350 K$ is the final temperature

Solving for V2, we find the final volume:

$V_2=\frac{V_1 T_2}{T_1}=\frac{(42)(350)}{184}=80 mL$

30)

For this problem, we can use the equation of state for ideal gases, which can be written as

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where in this problem:

$p_1 = 102.3 kPa$ is the initial pressure

$V_1=1975 L$ is the initial volume

$T_1=25^{\circ}C+273=298 K$ is the initial temperature

$p_2=21.5 kPa$ is the final pressure

$T_2=12^{\circ}C+273=285 K$ is the final temperature

And solving for V2, we find the final volume of the balloon:

$V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(102.3)(1975)(285)}{(21.5)(298)}=8987 L$

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A molecule of water consists of two atoms hydrogen bond with an atom of oxygen ( $H_2 O$ ) in a covalent bond.

While the molecul of water is overall neutral, due to the higher electronegativity of the oxygen atom, electrons are slightly shifted towards the oxygen atom; as a result, there is a slightly positive charge on the hydrogen side, and a slightly negative charge on the oxygen side (so, the molecules is said to be polar).

As a consequence, molecules of water attract each other, forming the so-called "hydrogen bonds".

One direct consequence of the polarity of water is that ice floats on liquid water.

Normally, for every substance on Earth, the solid state is more dense than the liquid state. However, this is not true for water, because ice is less dense than liquid water.

This is due to the polarity of water. In fact, when the temperature of water is decreased to freezing point and water becomes ice, the hydrogen bondings "force" the molecules to arrange in a lattice structure, so that the molecules become more spaced when they turn into solid state. As a result, ice occupies more volume than water, and therefore it is less dense, being able to float on water.

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