41.38 % Mg
55.17 % O
3.45 % H
Explanation:
What is the percent composition of magnesium hydroxide Mg(OH)₂?
To find the percent composition we follow the next algorithm.
First we calculate the molar mass of Mg(OH)₂:
molar mass of Mg(OH)₂ = molar mass of Mg × 1 + molar mass of O × 2 + molar mass of H × 2
molar mass of Mg(OH)₂ = 24 × 1 + 16 × 2 + 1 × 2 = 58 g/mole
Now we devise the next reasoning:
if in 58 g of Mg(OH)₂ there are 24 g of Mg, 32 g of O and 2 g of H
then in 100 g of Mg(OH)₂ there are X g of Mg, Y g of O and Z g of H
X = (100 × 24) / 58 = 41.38 % Mg
X = (100 × 32) / 58 = 55.17 % O
X = (100 × 2) / 58 = 3.45 % H
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percent composition
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The Correct answer is Carbon dioxide CO2.
The Explanation:
according to the diagram:
we can see that the arrows which going to A are from:
1- respirition of the tree . and the CO2 gas is the result and produced gas from respiration.
2- The Combustion went from Home : and the CO2 gas is also the produced gas from the combustion process.
∴ (A) refers to CO2 (carbon dioxide) gas.
Add all weighs together
calculate carbons weigh
(carb weigh/total)×100
Moles of MgS2O3 = 223/molar mass of MgS2O3
= 223/136.42
= 1.634 moles.
Hope this helps!
