Answer:
The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>
Explanation:
Given: The base dissociation constant: = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M
Also, water dissociation constant: = 1 × 10⁻¹⁴
<em><u>The acid dissociation constant </u></em>()<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>
<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>
Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+
Initial: 0.1 M x x
Change: -x +x +x
Equilibrium: 0.1 - x x x
<u>The acid dissociation constant: </u>
<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>
Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44
<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>
Answer:
Formula weight of H₂O molecule is 18.02 amu.
Explanation:
Given data:
Formula weight of H₂O = ?
Atomic mass of H = 1.008 amu
Atomic mass of O = 16.00 amu
Solution:
Formula weight:
"It is the sum of all the atomic weight of atoms present in given formula"
Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu
Formula weight of H₂O = 18.02 amu
Thus, formula weight of H₂O molecule is 18.02 amu.
(2) a base because they accept H+ ions. NH3 is the conjugate base of NH4+.
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
(119 g H2O) / (18.01532 g H2O/mol) x (1 mol Pb / 2 mol H2O) x (207.21 g Pb/mol) = 684 g Pb
Explanation: