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: The number of neutrons will increase as we move from left to right in a periodic table.
Explanation:
Atomic number is equal to the number of protons.
Mass number is the sum of number of neutrons and number of protons.
As we move from left to right, both the atomic number and mass number increases.
For example: As we move from Lithium to berrylium to boron to carbon to nitrogen to oxygen to fluorine to neon , the number of neutrons increase from 4 to 5 to 6 to 6 to 7 to 8 to 10 to 10.
Thus the number of neutrons will also increase as we move from left to right in a periodic table.
Answer:
open system
Coffee held in a cup is an open system because it can exchange matter water vapors and energy heat with the surroundings.
Answer:
pH = 6.8124
Explanation:
We know pH decreases with increase in temperature.
At room temperature i.e. 25⁰c pH of pure water is equal to 7
We know
Kw = [H⁺][OH⁻]...............(1)
where Kw = water dissociation constant
At equilibrium [H⁺] = [OH⁻]
So at 37⁰c i.e body temperature Kw = 2.4 × 10⁻¹⁴
From equation (1)
[H⁺]² = 2.4 × 10⁻¹⁴
[H⁺] = √2.4 × 10⁻¹⁴
[H⁺] = 1.54 × 10⁻⁷
pH = - log[H⁺]
= - log{1.54 × 10⁻⁷}
= 6.812
