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Answer:
CH3OH + 02 ----> C02 + H20
balanced equation -
CH3OH + 3/202 ----> C02 + 2H20
Use exactly the same process as the one used on another question of yours I answered.
Explanation:
First Reaction;
Ca + ZnCl2 --> CaCl2 + Zn
Oxidized Reactant: Ca. There is increase in oxidation number from 0 to +2
Reduced Reactant: Zn. There is decrease in oxidation number form +2 to 0
Second Reaction:
FeI2 + Mg --> Fe + MgI2
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Fe. There is decrease in oxidation number form +2 to 0
Third Reaction;
Mg + 2AgNO3 --> Mg(NO3)2 + Ag
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Ag. There is decrease in oxidation number form +1 to 0
Answer:
n₂ =1.4 mol
Explanation:
Given data:
Mass of nitrogen = 2 g
Initial Volume occupy by nitrogen = 1.25 L
Final volume occupy by nitrogen = 25.0 L
Final number of moles = ?
Solution;
Formula:
V₁ / n₁ = V₂ / n₂
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2 g/ 28 g/mol
Number of moles = 0.07 mol
Now we will put the values in formula:
V₁ / n₁ = V₂ / n₂
n₂ = V₂× n₁ /V₁
n₂ = 25 L × 0.07 mol / 1.25 L
n₂ = 1.75 L. mol / 1.25 L
n₂ =1.4 mol
