Answer:
1.
Since both components of these solutions have the same molar mass, mole fractions would be the same as mass fractions.
0.110 atm = (2/3)(Pi) + (1/3)(Pn) [1]
0.089 atm = (1/3)(Pi) + (2/3)(Pn) [2]
2*[1] - [2]:
(2)(0.110) - 0.089 atm = Pi
Pi = 0.131 atm
2*[2] - [1]:
(2)(0.089) - 0.110 atm = Pn
Pn = 0.068 atm
2.
The hydroxyl (-OH) group on the end of a longer 1-propanol molecule makes it more polar than IPA. It follows that the intermolecular forces between 1-propanol are stronger than those of IPA and thus the vapor pressure of 1-propanol should be lower than IPA.
Explanation:
When the balanced equation for this reaction is:
2Fe + 3H2O → Fe2O3 + 3H2
and according to the vapour pressure formula:
PV= nRT
when we have P is the vapor pressure of H2O= 0.121 atm
and V is the volume of H2O = 4.5 L
and T in Kelvin = 52.5 +273 = 325.5 K
R= 0.08205 atm-L/g mol-K
So we can get n H2O
So, by substitution:
n H2O = PV/RT
= (0.121*4.5)/(0.08205 * 325.5) = 0.02038 gmol
n Fe2O3 = 0.02038 * (1Fe2O3/ 3H2O) = 0.00679 gmol
Note: we get (1FeO3/3H2O) ratio from the balanced equation.
we can get the Mass of Fe2O3 from this formula:
Mass = number of moles * molecular weight
when we have a molecular weight of Fe2O3 = 159.7
= 0.00679 * 159.7 = 1.084 g
∴ 1.084 gm of Fe2O3 will produced
Answer:
0.88 g
Explanation:
Using ideal gas equation to calculate the moles of chlorine gas produced as:-

where,
P = pressure of the gas = 805 Torr
V = Volume of the gas = 235 mL = 0.235 L
T = Temperature of the gas = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles of chlorine gas = ?
Putting values in above equation, we get:

According to the reaction:-

1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.
So,
0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.
Moles of
= 0.01017 moles
Molar mass of
= 86.93685 g/mol
So,

Applying values, we get that:-

<u>0.88 g of
should be added to excess HCl (aq) to obtain 235 mL of
at 25 degrees C and 805 Torr.</u>