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The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
Answer:
1.02mole
Explanation:
The reaction equation is given as:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Given:
Mass of H₂SO₄ = 50g
Unknown:
Number of moles of NaOH = ?
Solution:
To solve this problem, we first find the number of moles of the acid given;
Number of moles =
Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 98g/mol
Now;
Number of moles = = 0.51mole
From the balanced reaction equation:
1 mole of H₂SO₄ will be neutralized by 2 mole of NaOH
0.51 mole of H₂SO₄ will be neutralized by 2 x 0.51 = 1.02mole of NaOH