Answer:
You will have 19.9L of Cl2
Explanation:
We can solve this question using:
PV = nRT; V = nRT/P
<em>Where V is the volume of the gas</em>
<em>n the moles of Cl2</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is 273.15K assuming STP conditions</em>
<em>P is 1atm at STP</em>
The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:
63g * (1mol / 70.906g) = 0.8885 moles
Replacing:
V = 0.8885mol*0.082atmL/molK*273.15K/1atm
V = You will have 19.9L of Cl2
Answer:
Explanation:
From the given information:
TO start with the molarity of the solution:
= 0.601 mol/kg
= 0.601 m
At the freezing point, the depression of the solution is
Using the depression in freezing point, the molar depression constant of the solvent
The freezing point of the solution
The molality of the solution is:
Molar depression constant of solvent X,
Hence, using the elevation in boiling point;
the Vant'Hoff factor
Answer is: 4.45 grams of methane gas <span>need to be combusted</span>.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V =
n·R·T.<span>
p = 1.1 atm.
T = 301 K.
V(H</span>₂O) <span>= 12.5 L.
R = 0,08206 L·atm/mol·K.
</span>n(H₂O) = <span>1.1 atm ·
12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
</span>n(H₂O) = 0.556 mol.
From chemical reaction: n(H₂O) : n(CH₄) = 2 : 1.
n(CH₄) = 0.556 mol ÷ 2 = 0.278 mol.
m(CH₄) = 0.278 mol · 16 g/mol.
m(CH₄) = 4.448 g.
Answer: 1.997 M
Explanation:
molarity = moles of solute/liters of solution or
first we have to find our moles of solute (mol), which you can find by dividing the mass of solute by molar mass of solute
mass of solute: 92 g
molar mass of solute: 46.08 g/mol
let's plug it in:
next, we plug it into our original equation:
Qualitative because it changed its colour not its quantity :)