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:
By visiting other households with cats.
Explanation:
This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.
Reduction. B is the correct answer
Answer:
328.1 K.
Explanation:
- To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT</em>.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in.
- If n is constant, and have two different values of (P, V and T):
<em>P₁V₁T₂ = P₂V₂T₁</em>
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P₁ = 1.0 atm (standard P), V₁ = 72.1 L, T₁ = 25°C + 273 = 298 K (standard T).
P₂ = 93.6 kPa = 0.924 atm, V₂ = 85.9 L, T₂ = ??? K.
<em>T₂ = P₂V₂T₁/P₁V₁ = </em>(0.924 atm)(85.9 L)(298 K)/(1.0 atm)(72.1 L) <em>= 328.1 K.</em>
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= 71.2 g