Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.
T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
Δn = 0,589 mol - 0,2356 mol = 0,3534 mol.
The problem can be solved using the following formula:
ΔTb = i Kb <em>m</em>
i = moles particles/moles solute
Kb = 0.512 °C/m
m = molality = moles solute/kg solvent
First we can solve for the molality of the solution:
75.0 g ZnCl₂ / 136.286 g/mol = 0.550 mol ZnCl₂
m = 0.550 mol/0.375 kg
m = 1.468 mol/kg
We can now solve for the change in temperature of the boiling point:
ΔTb = i Kb m
ΔTb = (3 mol particles/1 mol ZnCl₂) (0.512 °C/m) (1.468 m)
ΔTb = 2.25 °C
The boiling point of a solution is the initial boiling point plus the change in boiling point:
BP = 100 °C + 2.25 °C
BP = 102.25 °C
The solution will have a boiling point of 102.25 °C.
Answer:
we will take a 100g sample of this solution for our convenience
so , there is 15 g kBr in this 100g solution
we know that molality is the number if moles of solute / mass of solvent in kg
we need to find the number of moles in 15g kBr
no of moles = 15/119 s
moles = 0.126 moles/ 100g
multiplying both the numerator and the denominator by 10 to get 1 kg in denominator
= 1.26 moles / 1 kg
Hence, the molality is 1.26
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Answer:
b. 7.5 x 10^-3
Explanation:
To solve this problem we need to keep in mind the <em>definition of molarity</em>:
- Molarity = moles of solute / liters of solution
With the above information in mind it is possible to calculate the moles of solute, given the volume (10 mL) and concentration (0.75 M) of the solution:
- First we<u> convert 10 mL to L</u> ⇒ 10 mL / 1000 = 0.01 L
Then we <u>calculate the moles of AgNO₃</u>:
- moles of solute = Molarity * Liters of solution
- 0.01 L * 0.75 M = 7.5x10⁻³ mol AgNO₃
<em>One mole of AgNO₃ contains one mole of Ag⁺</em>, thus the number of Ag⁺ moles is also 7.5x10⁻³.
Answer:
I'm sorry to say but there is no statements in this question meaning no one will be able to answer them with out them.. what are the statements?
Explanation:
