Answer:
1. CO₂ → C + O₂
2. Fe₂ + O₂ → 2FeO
3. 2Al + 3CuO→ Al₂O₃ + 3Cu
Explanation:
1. 1 mol of CO₂ decomposes to 1 mol of C and 1 mol of oxygem
2. 1 mol of Fe₂ reacts with 1 mol of oxygen to produce 2 moles of iron (II) oxide
3. 2 moles of Al, reacts with 3 moles of cupper(II) oxide to produce 1 mol of aluminum dioxide and 3 moles of cupper
About 1 gram. 1 part of 50 is 2%
Answer:
4.90 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = ? M
V₁ = 125.0 mL
M₂ = 4.56 M
V₂ = 134.1 mL
Using the above formula , the molarity of acid , can be calculated as ,
M₁V₁ = M₂V₂
Substituting the respective values ,
M₁ * 125.0 mL = 4.56 M * 134.1 mL
M₁ = 4.90 M
The coefficients in a chemical equation represent the molar ratio of the substances.
For example, if an equation says 2H2 + O2 ⇒ 2H2O, it means
2 moles of H2 + 1 mol of O2 ⇒ 2 moles of H2O.
Answer:
The amount left after 49.2 years is 3mg.
Explanation:
Given data:
Half life of tritium = 12.3 years
Total mass pf tritium = 48.0 mg
Mass remain after 49.2 years = ?
Solution:
First of all we will calculate the number of half lives.
Number of half lives = T elapsed/ half life
Number of half lives = 49.2 years /12.3 years
Number of half lives = 4
Now we will calculate the amount left after 49.2 years.
At time zero 48.0 mg
At first half life = 48.0mg/2 = 24 mg
At second half life = 24mg/2 = 12 mg
At 3rd half life = 12 mg/2 = 6 mg
At 4th half life = 6mg/2 = 3mg
The amount left after 49.2 years is 3mg.
