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) :

4p, 3d, 3p, 3s, 2p, 2s and 1s orbitals may be occupied during de-excitation.<span />
Answer:
The last option:
- NH₃ (aq) + H⁺ (aq) → NH₄⁺ (aq)
Explanation:
1) Word equation
- Aqueous ammonia + nitric acid → aqueous ammonium nitrate
2) Chemical (molecular) equation
- NH₃ (aq) + HNO₃ (aq) → NH₄ NO₃
3) Ionization reactions
Write the dissociation of the soluble ionic compounds:
4) Total ionic equation:
- NH₃ (aq) + H⁺ (aq) + NO₃⁻ (aq) → NH₄⁺ (aq) + NO₃⁻ (aq)
5) Net ionic equation
You must cancel the spectator ions, which are those ions that are repeated in both reactant and product sides, i.e. NO₃⁻. They are name spectator because they do not participate (change) during the reaction.
- NH₃ (aq) + H⁺ (aq) → NH₄⁺ (aq)
And that is the last choice of the list.
Answer:
The second row elements are called Actinidies . abbreviated as Ac
Explanation:
Answer:
The answer to your question is:
a) 0.023 mol of CdS
b) 0.12 mol of MoO3
c) 7.3 mol of AlPO4
Explanation:
a) CdS
MW = 112.4 + 32
= 144 g
144g ------------------ 1 mol
3.28g -------------- x
x = (3.28 x 1) / 144
x = 0.023 mol
b) MoO3
MW = 96 + 3(16) = 144 g
144g ------------------- 1 mol
17.5g ----------------- x
x = (17.5 x 1) / 144
x = 0.12 mol
c) AlPO4
MW = 27 + 31 + 4(16)
MW = 122 g
122 g ------------------ 1mol
890 g ------------------ x
x = (890 x 1) / 122
x = 7.3 mol of AlPO4