The Percent composition of MgO is 60.304 % and 39.696 % respectively.
<h3>
What percentage of the given chemical is there?</h3>
Mg + 1/2O₂ -------> MgO
We now want to know the percent composition of MgO in light of this reaction. Since the original reactants' and the product's mass are not provided, we can use the component's atomic weights to solve this problem.
Let's compute the molar mass of MgO since Mg has a molecular weight of 24.305 g/mol and O has a molecular weight of 15.999 g/mol:
molar mass of 
Now with this weight, let's see the percent composition of this compound:
Mg percentage =
%
O percentage = 
%
The Percent composition of MgO is 60.304 % and 39.696 % respectively.
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Answer:
Vapor pressure of water = 23.14torr
Explanation:
When you made a solution, vapor pressure decreases following Raoult's law:
<em>Where P is vapor pressure and X mole fraction</em>
As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).
1000g of H₂O are, in moles (Molar mass: 18.02g/mol):
1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.
As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:
Replacing in Raoult's law, pressure of water above the solution is:
<h3>Vapor pressure of water = 23.14torr</h3>
Hey there!:
Given the reaction:
NaBr ⇌ Na⁺ + Br⁻
↓ ↓ ↓
0.150M 0.150M 0.150M
AgBr ⇌ Ag⁺ + Br⁻
↓ ↓ ↓
x x 0.150M
Therefore:
Ksp = x * 0.150
x = ( 7.7 * 10⁻¹³ ) / 0.150
x = 5.1 * 10⁻¹²
Answer B
Hope that helps!
