Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition =
× 100
% composition =
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
Answer:
Magnetic fields, revolve, asteroids
Answer:
625 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 250 mL
Molarity of stock solution (M₁) = 5 M
Molarity of diluted solution (M₂) = 2 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5 × 250 = 2 × V₂
1250 = 2 × V₂
Divide both side by 2
V₂ = 1250 / 2
V₂ = 625 mL
Therefore, the volume of the diluted solution is 625 mL.