Explanation:
the correct empirical formula for C6H12O6 is CH2O...
hope it will help....
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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Answer:
The reaction in this experiment is termed an iodine clock reaction, because it is the molecular iodine (I2) that undergoes the sudden concentration change. When the iodine concentration increases, it reacts with the starch in the solution to form a complex, turning it a deep blue-black color.
