<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:
where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:
Hence, the experimental van't Hoff factor is 1.21
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Somewhat false
observations can be made of a model of the statue of liberty, say, or in real line
Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.
Solution :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 3 atm
= final pressure of gas = ?
= initial volume of gas = 1.40 L
= final volume of gas = 0.950 L
= initial temperature of gas =
= final temperature of gas =
Now put all the given values in the above equation, we get the final pressure of gas.
Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.