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:
1) <em>The correct answer is A. Collision</em>
2) A hot solvent helps a solid dissolve faster because an increase in <u><em>kinetic energy</em></u> that also increases the rate of collisions
Explanation:
When a solute is added into a solvent and stirred, the solute particles get distributed to all parts of the solvent as a result of stirring.
More collisions occur between the solute and the solvent due to stirring. This increases the rate of dissolving.
<em>When a solvent is heated, then the kinetic energy would increase and the atoms will collide with a much greater force. As a result, ore solute will be able to dissolve in the solvent. </em>
Explanation:
It <em><u>provides an objective, standardized approach to conducting experiments</u></em> and, in doing so, improves their results. By using a standardized approach in their investigations, scientists can feel confident that they will stick to the facts and limit the influence of personal, preconceived notions.
I hope this helps you out!
Answer:
182.156g
Explanation:
grams = 49/.269 = 182.156g needed