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.
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The mass of a 0.513 mol of Al2O3 is 52.33g.
HOW TO CALCULATE MASS:
The mass of a substance can be calculated by multiplying the molar mass of the substance by its number of moles. That is;
mass of Al2O3 = no. of moles of Al2O3 × molar mass of Al2O3
According to this question, there are 0.513 moles of Al2O3.
Mass of Al2O3 = 0.513 × 102
Mass of Al2O3 = 52.33g
Therefore, the mass of a 0.513 mol of Al2O3 is 52.33g.
Learn more about mass calculations at: brainly.com/question/8101390?referrer=searchResults
