Answer:
1.26 × 10^-8 M
Explanation:
We are given;
Number of moles of mercury (i) chloride as 0.000126 μmol
Volume is 100 mL
We are required to calculate the concentration of the solution.
We need to know that;
Concentration is also known as molarity is given by;
Molarity = Number of moles ÷ Volume
Number of moles = 1.26 × 10^-10 Moles
Volume = 0.01 L
Therefore;
Concentration = 1.26 × 10^-10 Moles ÷ 0.01 L
= 1.26 × 10^-8 M
Thus, the molarity of the solution is 1.26 × 10^-8 M
The correct answer is letter C
<h3>Answer:</h3>
162.43 g of FeCl₂
<h3>
Explanation:</h3>
Step 1: Calculate mass of Fe;
As,
Density = Mass ÷ Volume
Or,
Mass = Density × Volume
Where Volume is the volume of water displaced = 10.4 mL
Putting values,
Mass = 7.86 g.mL⁻¹ × 10.4 mL
Mass = 81.744 g of Fe
Step 2: Calculate amount of FeCl₂;
The balance chemical equation is as follow,
Fe + 2 HCl → FeCl₂ + H₂ ↑
According to this equation,
55.85 g (1 mol) Fe produced = 110.98 g (1 mol) of FeCl₂
So,
81.744 g Fe will produce = X g of FeCl₂
Solving for X,
X = (81.744 g × 110.98 g) ÷ 55.85 g
X = 162.43 g of FeCl₂
Answer:
%yield of NH₃ = 30%
Explanation:
Actual yield of NH₃ = 40.8g
Theoretical yield = ?
Equation of reaction
N₂ + 3H₂ → 2NH₃
Molar mass of NH₃ = 17g/mol
Molarmass of N = 14.00
2 molecules of N = 2 * 14.00 = 28g/mol
Number of moles = mass / molar mass
Mass = number of moles * molar mass
Mass = 1 * 28.00 = 28g of N₂ (the number of moles of N₂ from the equation is 1).
From the equation of reaction,
28g of N₂ produce (2 * 17)g of NH₃
28g of N₂ = 34g of NH₃
112g of N₂ = x g of NH₃
X = (112 * 34) / 28
X = 136g of NH₃
Theoretical yield = 136g of NH₃
% yield = (actual yield / theoretical yield) * 100
% yield = (40.8 / 136) * 100
% yield = 0.3 * 100
% yield = 30%
