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
The balanced chemical reaction is :
Number of moles of Na, 
.
Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.
So, number of moles of oxygen are :
So, amount of oxygen required is :
Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.
You don’t have any answer chooses listed but i’ll say F isn’t a molecule because it stands for fluorine which happens to be an element that only has one atom when a molecule is supposed to have two or more atoms.
Answer:
5 L
Explanation:
We'll begin by calculating the molarity of the CaCl₂ solution. This can be obtained as follow:
Mole of CaCl₂ = 0.5 mole
Volume = 2 L
Molarity =?
Molarity = mole /Volume
Molarity = 0.5 / 2
Molarity = 0.25 M
Finally, we shall determine the volume of the diluted solution. This can be obtained as follow:
Molarity of stock solution (M₁) = 0.25 M
Volume of stock solution (V₁) = 2 L
Molarity of diluted solution (M₂) = 0.1 M
Volume of diluted solution (V₂) =?
M₁V₁ = M₂V₂
0.25 × 2 = 0.1 × V₂
0.5 = 0.1 × V₂
Divide both side by 0.1
V₂ = 0.5 / 0.1
V₂ = 5 L
Thus the volume of the diluted solution is 5 L
The simplest level at which life may exist is a cell
