One mole of an ideal gas will occupy 22.4 liters of volume at STP.
If the only thing that changes is the pressure from 1.00 atm to 2.00 atm, then we know the initial volume, initial pressure, and final pressure. We can use Boyle's law.
Boyle's law: (initial pressure)*(initial volume) = (final pressure)*(final volume)
fill in the knowns
(1.00 atm)*(22.4 liters) = (2.00 atm)(final volume)
divide both sides by 2.00 atm
11.2 liters = final volume
<span>The form of emission that is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass numbers is the gamma ray.
Gamma rays have no electrical charge nor atomic mass. It is only used to balance in a nuclear reaction, but gamma ray is considered powerful.
So, letter C. is the answer.
Hope this helps!</span>
Answer:
The Henry's law constant for argon is
Explanation:
Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.
C = k*P
where C is the solubility, P the partial pressure and k is the Henry constant.
So, being the concentration
where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So
In this case:
- P=PAr= 1 atm
- V=VAr= 5.16*10⁻² L
- R=0.082
- T=25 °C=298 °K
Then:
Solving:
n= 2.11 *10⁻³ moles
So:
Using Henry's Law and being C=CAr and P
=PAr:
2.11*10⁻³ M= k* 1 atm
Solving:
You get:
<u><em>The Henry's law constant for argon is </em></u><u><em></em></u>
Lower temperature
Let's verify
- Pressure=P
- volume=V
- Temperature=T
As per Boyles law
As per Charles law
So
At higher altitudes lower the pressure so lower the temperature