The volume in liters occupied by 22.6 g of I₂ gas at STP is 1.99 L (answer A)
<u><em>calculation</em></u>
Step: find the moles of I₂
moles= mass÷ molar mass
from periodic table the molar mass of I₂ is 253.8 g/mol
moles = 22.6 g÷253.8 g/mol =0.089 moles
Step 2:find the volume of I₂ at STP
At STP 1 moles =22.4 L
0.089 moles= ? L
<em>by cross multiplication</em>
={ (0.089 moles x 22.4 L) /1 mole} = 1.99 L
<span>The atomic weight of 13C should be pretty close to 13.0. (If you have the exact mass, use it in the problem.) So,
9.00 g / 13.0 g/mol = 0.692 moles
Therefore, the answer should be 0.692 moles are in 9.00 g of 13C.</span>
Answer:
Explanation:
It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.
Then you can consider it to be 11018 "moles" of "kJ"
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32.00
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ
n/mol: 7280
1. Moles of O₂
The molar ratio is 25 mol O₂:11 018 kJ
2. Mass of O₂
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
