The answer here would be 14.9 %KCl and here is how I can explain why:
33.5g / 225.6g x 100% = 14.9%
<span>you are looking for % of KCl in the solution, you have to add the mass of the KCl and water to get the total mass of the solution
Hope this helps a lot
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Answer is: cobalt(II) hydroxide.
Ksp(lead(II) chromate) = [Pb²⁺][CrO₄²⁻]. [Pb²⁺] = √2,8·10⁻¹³ = 5,3·10⁻⁷ M.<span>
Ksp(cobalt(II) hydroxide) = </span>[Co²⁺][OH⁻]². [Co²⁺] = ∛1,3·10⁻¹⁵ = 1,1·10⁻⁵ M.
Ksp(cobalt(II) sulfide) = [Co²⁺][S²⁻]. [Co²⁺] = √5·10⁻²² = 2,23·10⁻¹¹ M.
Ksp(chromium(III) hydroxide) = [Cr³⁺][OH⁻]³. [Cr³⁺] = √(√1,6·10⁻³⁰) = 3,6·10⁻⁸.
Ksp(silver sulfide) = [Ag⁺]²[S²⁻].
Answer:
600,000
Explanation:
In order to convert meters or metre to millimeter or millimetre, you must multiply the meter/metre by 1000. (Some countries say meter and others say metre.) Hope it helps! <3
The answer is Single Replacement.
Answer:
Percent yield of reaction is<em> 150%.</em>
Explanation:
Given data:
Percent yield = ?
Actual yield of SO₃ = 586.0 g
Mass of SO₂ = 705.0 g
Mass of O₂ = 80.0 g
Solution:
Chemical equation:
2SO₂ + O₂ → 2SO₃
Number of moles of SO₂:
Number of moles = mass/ molar mass
Number of moles = 586.0 g/ 64.1 g/mol
Number of moles = 9.1 mol
Number of moles of O₂:
Number of moles = mass/ molar mass
Number of moles = 80.0 g/ 32g/mol
Number of moles = 2.5 mol
Now we will compare the mole of SO₃ with O₂ and SO₂.
SO₂ : SO₃
2 : 2
9.1 : 9.1
O₂ : SO₃
1 : 2
2.5 : 2×2.5 = 5
The number of moles of SO₃ produced by oxygen are less it will limiting reactant.
Theoretical yield of SO₃:
Mass = number of moles × molar mass
Mass = 5 mol × 80.1 g/mol
Mass = 400.5 g
Percent yield of reaction:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 586.0 g/ 400.5 g× 100
Percent yield = 1.5× 100
Percent yield = 150%
