The mass of sodium chloride at the two parts are mathematically given as
- m=10,688.18g
- mass of Nacl(m)=39.15g
<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>
Generally, the equation for ideal gas is mathematically given as
PV=nRT
Where the chemical equation is
F2 + 2NaCl → Cl2 + 2NaF
Therefore
1.50x15=m/M *(1.50*0.0821)
1-50 x 15=m/58.5 *(1.50*0.0821)
m=10,688.18g
Part 2
PV=m'/MRT
1*15=m'/58.5*0.0821*273
m'=39.15g
mass of Nacl(m)=m'=39.15g
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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.
An electron has a negative charge of one
An neutron has no charge (hence, neutral)
An proton has a positive charge of one
~
Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
