Answer:
(a) 
(b) 
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

Thus, since the final pressure is 3.60 bar, we can write:

The moles of helium could be computed via solver as:

Or algebraically:

In such a way, the volume of the compartment B is:

Finally, he mole fraction of He is:

Regards.
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
Answer:
Gelada baboons plays a significant role.
Explanation:
The role of gelada baboons in their ecosystem is very important because they aerate the soil for plants which is necessary for good plant growth. These gelada baboons also helps in controlling the population of predator in their ecosystem which is very essential for the stability and equilibrium of the ecosystem so gelada baboons has a good effect on both plants and animals in their ecosystem.
Answer: Empirical formula is 
Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.
<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Molar mass of Carbon = 12.0 g/mol
Molar mass of Hydrogen = 1.0 g/mol
Molar mass of Oxygen = 16.0 g/mol
Moles of Carbon = 
Moles of Hydrogen = 
Moles of Oxygen = 
<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value
Smallest number of moles = 13.76 moles



<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula
Empirical formula becomes = 