A and C, Igneous and Metamorphic
A chemical bond is the best classification.
Answer:
(i) Change in colour (ii) Change in temperature (iii) Formation of precipitate
Explanation:
(i) Change in colour: Reaction between lead nitrate solution and potassium iodide solution. Pb(NO3)2(aq)+2KI → PbI2(s)+2KNO3(aq) In this reaction, colour changes from colourless to yellow. (ii)Change in temperature: Action of dilute sulphuric acid on zinc. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2 In this reaction, heat is evolved (iii) Formation of precipitate: Action of barium chloride on sodium sulphate. BaCl2(aq) +Na2SO4(aq) → BaSO4(s) +2NaCl(aq) BaSO4(s)
Answer:
a) 1.71 × 10⁻³ M
b) 8.00 × 10⁻⁵ M
Explanation:
In order to calculate the solubility (S) of Pb(SCN)₂ we will use an ICE chart. We identify 3 stages (Initial, Change, Equilibrium) and complete each row with the concentration or change in the concentration.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product (Ksp) is:
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (2S)² = 4S³
S = 1.71 × 10⁻³ M
<em>b) Calculate the molar solubility of lead thiocyanate in 0.500 M KSCN.</em>
KSCN is a strong electrolyte that dissociates to give 0.500 M K⁺ and 0.500M SCN⁻.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0.500
C +S +2S
E S 0.500 + 2S
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (0.500 + 2S)²
In the term (0.500 + 2S)², 2S is negligible.
Ksp = 2.00 × 10⁻⁵ = S . (0.500)²
S = 8.00 × 10⁻⁵ M
