Answer:
1.47 atm
Explanation:
Step 1: Given data
- Initial volume (V₁): 32.4 L
- Initial pressure (P₁): 1 atm (standard pressure)
- Initial temperature (T₁): 273 K (standard temperature)
- Final volume (V₂): 28.4 L
- Final temperature (T₂): 352 K
Step 2: Calculate the final pressure of the gas
We can calculate the final pressure of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
P₂ = P₁ × V₁ × T₂ / T₁ × V₂
P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm
-19.15 degrees Celsius. 254k-273.15=-19.15
You have to subtract 273
Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr
Answer:
condensation is the answer