Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.
Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
Answer
D 160g
Explanation:
<u>Write the equation:</u>
Combustion reactions use oxygen and release water and heat, so
CH₃OH(g) + O₂(g) → CO₂(g) + H₂O(g)
Balance that:
2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)
<u>Find moles of carbon dioxide:</u>
We need to know the number of moles of CO₂. This rxn is at STP, so at STP one mole of gas = 22.4 liters.
112 L * 1 mol/22.4 L = <em>5 mol CO₂</em>
<u>Find moles of methanol:</u>
Based on the chemical equation, for every 2 mol methanol, there are 2 mol carbon dioxide. So for every 5 mol carbon dioxide, there are 5 mol methanol!
5 mol CO₂ = 5 mol CH₃OH
Molar mass of methanol: 12.01 + 3*1.008 + 16.00 + 1.008 = <em>32.04 g/mol</em>
Moles of methanol: 5 mol * 32.04 g/mol = 160.2 g methanol
≈ 160 mol methanol
Answer : The normality of the solution is, 30.006 N
Explanation :
Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.
Mathematical expression of normality is:
or,
First we have to calculate the equivalent weight of solute.
Molar mass of solute 
= 94.97 g/mole
Now we have to calculate the normality of solution.
Therefore, the normality of the solution is, 30.006 N
Answer:
Explanation:
mass ratio of oxygen and nitrogen in air at Miami
= 21 : 79
ratio of their moles
= 
( mol weight of oxygen is 32 and of nitrogen is 28 )
= .65625 : 2.8214
= 1 : 4.3
This ratio will also be maintained in the air of Denver though total pressure decreases there.
Partial pressure of oxygen in air at both the places
mole fraction of oxygen
= 
= .18868
partial pressure of oxygen at Denver
= .18868 x .83 x 760
= 119 mm.
