Answer: -
Concentration of PbI₂ = 1.5 x 10⁻³ M
PbI₂ dissociates in water as
PbI₂ ⇄ Pb²⁺ + 2 I⁻
So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.
Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =
= 1.5 x 10⁻³ x 2 M
= 3 x 10⁻³ M
PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.
[Pb²⁺] = 1.5 x 10⁻³ M
So solubility product for PbI₂
Ksp = [Pb²⁺] x [ I⁻]²
=1.5 x 10⁻³ x (3 x 10⁻³)²
= 4.5 x 10⁻⁹
Answer:
1.1 × 10²⁴ atoms Mg
General Formulas and Concepts:
<u>Atomic Structure</u>
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify</em>
[Given] 1.8 mol Mg
[Solve] atoms Mg
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.08396 × 10²⁴ atoms Mg ≈ 1.1 × 10²⁴ atoms Mg
The limiting reactant is chlorine (Cl2).
<u>Explanation</u>:
Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.
2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.
Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g
= 1.260 g moles of Al
Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g
= 1.10 g moles of Cl2
Dividing each reactant by it's coefficient in the balanced equation obtains:
1.260 moles Al / 2 = 0.63 g moles of Al
1.11 moles Cl2 / 3 = 0.36 g moles of Cl2
The reactant which produces a lesser amount of product is called as limiting reactant.
Here the Limiting reactant is Cl2.
A. hex- Normally with 6, the prefix is hex-. For example a polygon having 6 sides is known as hexagon.
Answer:
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