Answer:
6 mins
Explanation:
The time taken for Ar to effuse can be obtained as follow:
Time for Br₂ (t₁) = 12 mins
Molar mass of Br₂ (M₁) = 2 × 80 = 160 g/mol
Molar mass of Ar (M₂) = 40 g/mol
Time for Ar (t₂) =?
t₂/t₁= √(M₂/M₁)
t₂ / 12 = √(40/160)
Cross multiply
t₂ = 12 × √(40/160)
t₂ = 12 × 0.5
t₂ = 6 mins
Therefore, it will take 6 mins for the same amount of Ar to effused out.
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.
0.019 Hope this helps! :)
C2H6O + 3O2 —> 2CO2 + 3H2O
The correct answer is A, since it is the only sensible answer.
