Answer:
0.3097 moles of an nonionizing solute would need to be added.
Explanation:
Molal elevation constant = 
Normal boiling point of ethanol = 
Boiling of solution =
Moles of nonionizing solute = n
Mass of ethanol (solvent) = 47.84 g
Elevation boiling point:
n = 0.3097 mol
0.3097 moles of an nonionizing solute would need to be added.
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
Answer: 5 plates
Explanation:
Because you have 5 sandwiches total
Answer:
THE PARTIAL PRESSURE OF OXYGEN GAS IN THE CONTAINER IS 92.67kPa WHICH IS OPTION B.
Explanation:
To calculate the partial pressure of oxygen gas collected over water, we use
Ptotal = Poxygen + P water
It is worthy to note that when oxygen is collected over water, it is mixed with water vapor and the total pressure in the container will be the sum of the pressure exerted by the oxygen gas and that of the water vapor at that given temperature.
At 20 C, the vapor pressure of water as given in the question is 2.33 kPa.
Using the above formula,
Ptotal = Poxygen + P water
Substituting for Poxygen, we have;
Poxygen = Ptotal - P water vapor
P oxygen = 95 .00 kPa - 2.33 kPa
P oxygen = 92.67 kPa.
The partial pressure of oxygen gas in the container is hence, 92.67kPa.
