A sample of Xe takes 75 seconds to effuse out of a container. An unknown gas takes 37 seconds to effuse out of the identical container under identical conditions. What is the most likely identity of the unknown gas?

a. Br2
b. He
c. O2
d. Kr

The most likely unknown gas has been . Thus, option C is correct.

The two gas required different times to be diffused from the same container. Thus, the rate of diffusion has been proportional to the molar mass of the substance.

Thus, the ratio of the rate of two samples can be given by:

Rate can be defined as the amount diffuses per unit time. The equation in terms of time can be given as:

Let gas B be Xenon, having diffusion time 75 sec, and molar mass 131.3 g/mol

Molar mass of A = 32 g/mol.

The gas with a molar mass of 32 g/mol has been oxygen. Thus, the most likely unknown gas has been . Thus, option C is correct.

For more information about the diffusion of gases, refer to the link:

link

The molar mass of B is 32.0 g/mol

The unknown gas might be O2 (option C)

Explanation:

Step 1: Data given

A sample of Xe takes 75 seconds to effuse out of a container.

An unknown gas takes 37 seconds to effuse out of the identical container

Molar mass Xe = 131.3 g/mol

Step 2:

Rate A / rate B = √ (MB/MA)

rate = amount / time

The equation includes rate, but you are given times. Since rate is amount/time, we have to invert that side of the equation.

Time B/Time A = √(MB/MA)

⇒with time B = the time the unknown gas needed to effuse  = 37 seconds

⇒with time A = the time needed Xe needed to effuse = 75 seconds

⇒with MB = the molar mass of the unknown gas = TO BE DETERMINED

⇒ with MA = the molar mass of Xe = 131.3

37/75 = √ (MB/131.3)

MB = 32

The molar mass of B is 32.0 g/mol

The unknown gas might be O2

there all right good

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%