1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
Answer: Chyme
Explain: Chyme passes from the stomach to the small intestine. Further protein digestion takes place in the small intestine.
C. The number of Valence electrons,
Every atom tries to follow the Octet rule i.e To have 8 electrons in its Valence shell.
Every atom tries to accomodate 8 electrons in its Valence shell to stabilize themselves, Metals usually have 1-3 eletrons in their Valence shell which they donate to non metals so their Valnce shell has 8 electroons, (The previous will now be the Valence shell and it will be full)
Similarly Non metals have 4-7 electrons, they accept electrons from metals so they can have 8 electrons in their Valence shell.
Noble gases already have 8 electrons in their Valance shell, so they do not react and stable.
Hope it helps :)
Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is 
.
Explanation:
When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.
For example, chemical equation for oxidation of methane is as follows.
Number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
To balance this equation, multiply 
by 2 on reactant side. Also, multiply 
by 2 on product side. Hence, the equation can be rewritten as follows.
Now, the number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.
Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is .
