I believe "h" stands for hecta-, which means "one hundred."
I will convert from g to hg (divide by 100), then from mL to L (multiply by 1000).
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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I believe the statement is true. The heat transfer involved here is called conduction. It is one type of heat transfer where it caused by the collisions of the particles in a certain body. As the you add heat to the object, the kinetic energy of the molecules increases therefore more collisions could happen which would mean energy is dissipated or transferred from molecule to molecule.<span />
