All categories

3 years ago

Find the ratio of the effusion rate of hydrogen gas to the effusion rate of krypton gas.

2 answers:

6 0

This problem could be solved through the Graham’s law of effusion (also known as law of diffusion). This law states that the ratio of the effusion rate of the first gas and effusion rate of the second gas is equivalent to the square root of the ratio of its molar mass. Thus the answer would be 0.1098.

ElenaW [278]3 years ago

3 0

Answer:

4.243

Explanation:

The effusion is the movement of the molecules for a hole by a difference of pressure. The effusion rate is inversely proportional to the molar mass of the gas (M), and the ratio of the rates (R) of two gases can be found by Graham' Law:

R1/R2 = (√M2)/(√M1)

Let's call hydrogen gas as 1 (M1 = 2 g/mol), and krypton gas as 2 (M2 = 36 g/mol):

R1/R2 = (√36)/(√2)

R1/R2 = 4.243

You might be interested in

Which of the following is an example of an observation?

Bess [88]

THE ANSWER TO THE QUESTION IS A

8 0

3 years ago

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl

4vir4ik [10]

Well, we need to find the ratio of Al to the other reactant.

Al:HCl = 1:3

--> this means that for every 1 Al used, you have to use 3 HCl.

6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.

The ratio of HCl:AlCl = 3:1

13/3 = 4.3333...

The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.

Hope this helps!!! :)

6 0

3 years ago

Describe Rutherford's contribution to the atomic model.

Nady [450]

Answer:

Rutherford's experiment, also known as

$\alpha - scattering \: experiment$

supports the existence of neutrons and the nucleus.

Explanation:

In the above diagram, Rutherford was trying to explain his contributions using thin foils of gold and other metals as targets for alpha particles from a radioactive source.

He observed that the majority of particles penetrated the foil either undeflected or with only a slight deflection. But, every now and then an alpha particle was scattered(or deflected) at a large angle..

According to Rutherford, most of the atoms must be empty space. This explains why the majority of alpha particles passed through through the gold foil with little or no deflection. The atoms positive charges, Rutherford proposed are all concentrated in the Nucleus, which is a dense central core within the atom.

Whenever an alpha particle came close to a nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an alpha particle coming towards a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the Nucleus are called Protons.

I hope you find this useful... Have a lovely day. 

4 0

3 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

8 0

2 years ago

Gas laws describe and predict the behavior of gases without attempting to explain why they happen

Nutka1998 [239]

The gas laws describe and predict the behavior of gases with an explanation and experimental data

So the given statement is False.

2) The volume of gas can be calculated based on Avagadro's law

It states that the volume of a gas is directly proportional or varies with the moles of the gas. Higher the moles more the volume, condition is the pressure and temperature are constants in the two conditions

Thus as here the pressure and temperature of nitrogen gas is kept constant

V α moles

$\frac{V1}{n1}=\frac{V2}{n2}$

Where

V1 = 6 l

n1 = 0.50 mol

V2 = ?

n2 = 0.75 mol

On putting values

V2 = 6 X 0.75 / 0.5 = 9 L

so resulting volume of the gas will be 9L

5 0

3 years ago

Read 2 more answers