Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
Answer:
5.00 mol Mg
10.0 mol Cl
40.0 mol O
Explanation:
Step 1: Given data
Moles of Mg(ClO₄)₂: 5.00 mol
Step 2: Calculate the number of moles of Mg
The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.
5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg
Step 3: Calculate the number of moles of Cl
The molar ratio of Mg(ClO₄)₂ to Cl is 1:2.
5.00 mol Mg(ClO₄)₂ × 2 mol Cl/1 mol Mg(ClO₄)₂ = 10.0 mol Cl
Step 4: Calculate the number of moles of O
The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.
5.00 mol Mg(ClO₄)₂ × 8 mol O/1 mol Mg(ClO₄)₂ = 40.0 mol O
A the type of soil the tulip receive- with or without fertilizer
1 is B (Just remember to have the same number of atoms on both sides)
2 is B (A precipitate is a solid forming from 2 liquids)
