Dissociation of para-bromobenzoic acid can be represented as:
4-BrC₆H₄COOH----->4-BrC₆H₄COO⁻ + H⁺
Dissociation constant of this acid can be calculated as:
Ka={[BrC₆H₄COO⁻][H⁺]}/[4-BrC₆H₄COOH]
as[4-BrC₆H₄COOH=Concentration of para- bromobenzoic acid=0.35 M
And as 1.69% of this acid dissociates to form [BrC₆H₄COO⁻] and [H⁺], so amount of these ions formed will be:
[BrC₆H₄COO⁻]=[H⁺]=1.69×0.35=0.59
So Now Ka=(0.59×0.59)/0.35
=0.99.
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u /> = 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
We are requested to decide the warmth radiated in the response. Warmth emitted is resolved through the condition triangleH = mCp triangleT where m is the mass, Cp is the particular warmth and triangleT is the temperature contrast. Substituting, triangleH = 120 g (4.184 j/g •°c) (29.2°c). Heat given off is equal to 14.66 kJ or 14.7 rounded