Answer:
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Explanation:
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Answer:
2.7 °C.kg/mol
Explanation:
Step 1: Calculate the freezing point depression (ΔT)
The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:
ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C
Step 2: Calculate the molality of the solution (b)
We will use the following expression.
b = mass of solute / molar mass of solute × kilograms of solvent
b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg
Step 3: Calculate the molal freezing point depression constant Kf of X
Freezing point depression is a colligative property. It can be calculated using the following expression.
ΔT = Kf × b
Kf = ΔT / b
Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol
Answer:
81.5 L
Explanation:
We can use the combined gas law equation that gives the relationship among pressure, temperature and volume of gases for a fixed amount of gas.
P1V1 / T1 = P2V2 / T2
where P1 - pressure, V1 - volume and T1 - temperature at the first instance
P2 - pressure, V2 - volume and T2 - temperature at the second instance
substituting the values in the equation
1240 Torr x 47.2 L / 298 K = 730 Torr x V2 / 303 K
V2 = 81.5 L
the new volume the gas would occupy when the conditions have changed is 81.5 L
Answer & Explanation:
The molar mass of calcium chloride is 110.98 g/mol. We can use this information to solve this problem. We can set up our equation like this..
Multiply straight across on the top and straight across on the bottom.
Now divide.
So, there are 3.00 moles of calcium chloride contained in a 33 gram sample which is answer choice D.
