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:
Rewrite the following sentence in reported speech
The chair person said,"A hearty welcome to all present!"
N is newtons so it would be false
Your answer to this question would be B. packaging waste.
hope this helps!
The volume increases to 1.009 L.
<em>V</em>= <em>V</em>_0 +βΔ<em>T</em>
The thermal expansion coefficient (β) of water changes with temperature, so we must calculate the volume change over small (10 °C) intervals.
20 °C to 30 °C: <em>V</em> = 1 L + 0.000 207 L·°C^(-1) × 10 °C = 1.002 07 L
30 °C to 40 °C: <em>V</em> = 1.002 07 L + 0.000 303 L·°C^(-1)] × 10 °C = 1.005 10 L
40 °C to 50 °C: <em>V</em> = 1.005 10 L + 0.000 385 L·°C^(-1)] × 10 °C = 1.008 95 L
The volume increases by about 9 mL when the temperature increases from 20 °C to 50 °C.
