Answer:
1.00 M
Explanation:
Sn^2+ reacts with KMNO4 as follows;
5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)
The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L
= 0.0061 moles
If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-
x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-
x= 5 × 0.0061/2
x= 0.015 moles
Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L
number of moles = concentration × volume
Concentration = number of moles/volume
Concentration= 0.015 moles/0.015 L
Concentration = 1 M
The actual yield is 43 g Cl₂.
The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.
∴ The <em>theoretical yield</em> is 60.25 g Cl₂.
% yield = actual yield/theoretical yield × 100 %
Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g
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When 1. 0 l of 0. 00010 m NaOH and 1. 0 l of 0. 0014 m mgso4 are mixed, there will be no precipitate formed.
<h3>What is a precipitate?</h3>
The precipitate is the solid concentration of a substance that is collected over a solution.
First, we determine the concentration of magnesium and hydroxide
(Mg2+) = 7.00 × 10⁻⁴
(OH−) = 5.00 × 10⁻⁵
Now, we calculate the solubility quotient
Qc = (Mg2+) (OH−) ²
Qc = 7.00 × 10⁻⁴ x (5.00 × 10⁻⁵)²
Qc = 1.75 x 10⁻¹²
The solubility product of the magnesium hydroxide is 1.80 x 10⁻¹¹ which is more than the solubility quotient. Thus, there will be no precipitate form.
Thus, there will be no precipitate formed because the solubility quotient we calculated is less than the solubility product.
To learn more about precipitate, refer to the below link:
brainly.com/question/16950193
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