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.
Milk is a complex colloidal system.
Answer:
The correct answer is "The coffee in the jug has more thermal energy than the coffee in the cup".
Explanation:
First I had to look for the problem to know the possible answers.
In this case, the coffee jug has a large amount of coffee at the same temperature. If we analyze that the decanter and the coffee are at the same temperature, we have a homogeneous thermal system. The cup is at room temperature, so by pouring coffee into it, the temperature of the coffee decreases to balance with the temperature of the cup. At this moment, the temperature of the cup-cafe system is lower than the jug-cafe system.
Thermal energy is the part of the internal energy of an equilibrated thermodynamic system that is proportional to its absolute temperature and increases or decreases by energy transfer.
In this way, we can ensure that the thermal energy of the cup-cafe system is lower than that of the jug-cafe system.
Have a nice day!
Answer:
pH=8.676
Explanation:
Given:
0.75 M 
0.20 M 
The objective is to calculate the pH of the buffer using the kb for 
Formula used:
![pOH=pka+log\frac{[salt]}{[base]}\\](https://tex.z-dn.net/?f=pOH%3Dpka%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%5C%5C)
pH=14-pOH
Solution:
On substituting salt=0.75 and base=0.20 in the formula
pH=14-pOH
On substituting the pOH value in the above expression,
pH=14-5.324
Therefore,
pH=8.676
