Answer:
C₃H₆O₃
Explanation:
Data:
EF = CH₂O
MM = 90. g/mol
Calculations:
EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ

MF = (CH₂O)₃ = C₃H₆O₃
Answer:
V₁ = 10 mL
Explanation:
Given data:
Initial volume of HCl = ?
Initial molarity = 3.0 M
Final molarity = 0.10 M
Final volume = 300.0 mL
Solution:
Formula:
M₁V₁ = M₂V₂
M₁ = Initial molarity
V₁ = Initial volume of HCl
M₂ =Final molarity
V₂ = Final volume
Now we will put the values.
3.0 M ×V₁ = 0.10 M×300.0 mL
3.0 M ×V₁ = 30 M.mL
V₁ = 30 M.mL /3.0 M
V₁ = 10 mL
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 =
=74.8 %.
C) Nucleus...... Hope it helps, Have a nice day :)