Density/Earth’s gravitational pull.
Answer:
1.9 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 1.5 atm
- Initial volume (V₁): 3.0 L
- Initial temperature (T₁): 293 K
- Final pressure (P₂): 2.5 atm
- Final temperature (T₂): 303 K
Step 2: Calculate the final volume of the gas
If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂ / T₁ × P₂
V₂ = 1.5 atm × 3.0 L × 303 K / 293 K × 2.5 atm = 1.9 L
Answer:
75.15 mol.
Explanation:
- Firstly, we need to write the balanced equation of the reaction:
<em>Fe₂O₃ + 3CO → 2Fe + 3CO₂.</em>
It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.
∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.
- we need to calculate the no. of moles of (4000 g) of Fe₂O₃:
<em>no. of moles of Fe₂O₃ = mass/molar mass</em> = (4000 g)/(159.69 g/mol) = <em>25.05 mol.</em>
<u>Using cross multiplication:</u>
1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,
∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.
<em>∴ The no. of moles of CO needed</em> = (3.0 mol)(25.05 mol)/(1.0 mol) =<em> 75.15 mol.</em>
Answer:
This is a coal combustion process and we will assume
Inlet coal amount = 100kg
It means that there are
15kg of H2O, 2kg of Sulphur and 83kg of Carbon
Now to find the mole fraction of SO2(g) in the exhaust?
Molar mass of S = 32kg/kmol
Initial moles n of S = 2/32 = 0.0625kmols
Reaction: S + O₂ = SO₂
That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂
Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust
The mole fraction of SO2(g) in the exhaust=0.0625kmols
Explanation:
