The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ
= i
m
where:
Δ = change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration = 
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ = im we get,
Δ = 2 × 1.86 × 0.1219
Δ = 0.4536°C
So, Δ of solution is,
Δ
= 0.00°C - 0.45°C
Δ = - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
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A = Z + n
A = 12 + 10
A = 22
Answer C
hope this helps!
The answer is predict, because it fits, and because someone can predict the outcome of something.
Answer:
Subtract water vapor pressure from total pressure to get partial pressure of gas A: PA=1.03 atm- 1 atm=0.03 atm.
What is the total pressure of the gases at 298 K?
98.8 kPa
A sample of nitrogen gas is bubbled through water at 298 K and the volume collected is 250 mL. The total pressure of the gas, which is saturated with water vapour, is found to be 98.8 kPa at 298 K.
The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: Ptotal=P1+P2+… +Pn. + P n . The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.
How do you find the partial pressure of water in air?
e is vapor pressure Rv = R∗/Mv = 461.5Jkg−1K−1 and Mv = 18.01gmol−1, ϵ = Mv/Md = 0.622. The vapor pressure is the partial pressure of the water vapor. where es is in Pascals and T is in Celsius.
ExpHow do you find the pressure of h2?
For the high pressures in which hydrogen gas is often stored, the van der Waals equation can be used. It is P+a(n/V)^2=nRT. For diatomic hydrogen gas, a=0.244atm L^2/mol^2 and b=0.0266L/mol.lanation:
We can express the rate equation in this form:
-r = k A^n B^m
where -r is the rate
k is the rate constant,
A is the concentration of CH3Cl
n is the order with respect to CH3Cl
B is the concentration of H2O
m is the order with respect to H2O
We can solve this by trial and error or by calculus. The first method is easier. The rate constant does not depend on the concentration of the reactant. Assume values of n and m and solve for k in each experiment. The only option that gives really close values of k in each experiment is:
<span>C. CH3Cl: firstorder H2O: second order
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