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STALIN [3.7K]
2 years ago
12

If two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is

Chemistry
1 answer:
Fed [463]2 years ago
3 0

If the gases are at the same temperature and pressure, the ratio of their effusion rates is directly proportional to the ratio of the square roots of their molar masses:

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

Where

  • R₁ and R₂ are the rates of the two gas
  • M₁ and M₂ are the molar masses of the two gas

From the Graham's law equation, we can see that the ratio of the rates of effusion of the two gases is directly proportional to the square root of their molar masses

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

#SPJ1

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Answer:

pulmonary vein

Explanation:

The pulmonary vein is the only vein in the body that carries oxygen rich blood.

4 0
2 years ago
Calculate the molality of 75.0 grams of MgCl2 (molar mass=95.21 g/mol) dissolved in 500.0 g of solvent.
nordsb [41]

<u>Answer:</u> The molality of magnesium chloride is 1.58 m

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (magnesium chloride) = 75.0

M_{solute} = Molar mass of solute (magnesium chloride) = 95.21 g/mol  

W_{solvent} = Mass of solvent = 500.0 g

Putting values in above equation, we get:

\text{Molality of }MgCl_2=\frac{75.0\times 1000}{95.21\times 500.0}\\\\\text{Molality of }MgCl_2=1.58m

Hence, the molality of magnesium chloride is 1.58 m

5 0
3 years ago
Read 2 more answers
An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi
Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow  = 713.51 N/m

3 0
2 years ago
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7 0
3 years ago
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What are some examples of properties scientists use to describe matter?
AveGali [126]

Answer:

Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.

4 0
2 years ago
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