Answer:
pH = 10.505
Explanation:
Molar mass of Amphetamine ( C9H13N) = 135 g/mol
Given that the concentration of Amphetamine = 225 mg/L
mass of Amphetamine in one Liter =
= 0.225 g
Number of moles of Amphetamine in one liter =
= 0.001667 mol
∴ molarity = 0.0017 M
C₉H₁₃N + H₂O --------> C₉H₁₃NH⁺ + OH⁻
I(M) 0.001667 M 0 0
C(M) -x x x
E(M) 0.001667 - x x x
Pkb = -log Kb = 4.2
∴ Kb = 6.309 x 10⁻⁵
Kb = 6.309 x 10⁻⁵
Equilibrium constant = [C₉H₁₃NH⁺][OH⁻]/ [C₉H₁₃N]
6.309 x 10⁻⁵ = x² / 0.001667-x
where 0.001667 -x ≅ 0.001667
Then;
x² = 6.309 x 10⁻⁵ × 0.001667
x² = 1.0517103 × 10⁻⁷
x = 
x = 0.00032 M
x = [OH-] = 0.00032 M
∴ pOH = -log [OH-]
pOH = -log (0.00032)
pOH =3.495
pH = 14 - 3.495
= 10.505
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Answer:
Polar covalent bond because of difference in electronegativities
Explanation:
The mass percentage of each element in the milk of magnesia is, 41.68% Mg, 54.86% Oxygen, and 3.46% Hydrogen elements
Give the total mass of a sample of milk of magnesia Mg(OH)₂ as 58.33 g
We are to calculate the percentage by mass of each element

For Magnesium:
Mass of magnesium = 24.31g

For oxygen element:
Mass of oxygen = 32.00g

For the hydrogen element
Mass of hydrogen = 2.02g

Hence the mass percentage of each element in the milk of magnesia is, 41.68% Mg, 54.86% Oxygen, and 3.46% Hydrogen elements.
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Answer:
6.69%
Explanation:
Given that:
Mass of the fertilizer = 0.568 g
The mass of HCl used in titration (45.2 mL of 0.192 M)
= 
= 0.313 g HCl
The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)
= 
= 0.0058919 mole of NaOH
From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process
Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g
= 0.2151 g HCl
From above ; the total amount of HCl used = 0.313 g
The total amount that is used for complete neutralization = 0.2151 g
∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g
= 0.0979 g
We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g
Now; the amount of HCl neutralized by 0.0979 HCl = 
= 0.0456 g
Therefore, the mass of nitrogen present in the fertilizer is:
= 
= 0.038 g
∴ Mass percentage of Nitrogen in the fertilizer =
%
= 6.69%