0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
Given:
n = 0.0021 mol
M = 44 gram/mol
Required: m or the mass of the molecule
Solution:
M = m/n
Rearranging the expression,
m = M x n
m = 44 g/mol x 0.0021 mol
Evaluating the expression and cancelling units, we obtain,
m = 0.0924 grams
M is the molar mass. This value is very useful in chemistry calculations especially in calculating the amounts of reactants and products for a chemical reaction.
<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
<h3><u>Answer;</u></h3>
Empirical formula = C₂H₃O
Molecular formula = C₁₄H₂₁O₇
<h3><u>Explanation</u>;</h3>
Empirical formula
Moles of;
Carbon = 55.8 /12 = 4.65 moles
Hydrogen = 7.04/ 1 = 7.04 moles
Oxygen = 37.16/ 16 = 2.3225 moles
We then get the mole ratio;
4.65/2.3225 = 2.0
7.04/2.3225 = 3.0
2.3225/2.3225 = 1.0
Therefore;
The empirical formula = <u>C₂H₃O</u>
Molecular formula;
(C2H3O)n = 301.35 g
(12 ×2 + 3× 1 + 16×1)n = 301.35
43n = 301.35
n = 7
Therefore;
Molecular formula = (C2H3O)7
<u> = C₁₄H₂₁O₇</u>
