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
Explanation:
option no 4 is correct answer
I hope is helpful
Answer:
I think its D
Explanation:
It cant be B or C bc the solute particles arent changing. And its not A because the concentration isnt constant
<span>The following acids do NOT ionize completely in solution because they are not strong acids are as follows:
a HBr
b HI
c H2SO3
d H3N
e HNO2
f HF
</span>
According to Henderson–Hasselbalch Equation,
pH = pKa + log [Lactate] / [Lactic Acid]
As,
Ka of Lactic Acid = 1.38 × 10⁻⁴
pKa = -log Ka
pKa = -log 1.38 × 10⁻⁴
pKa = 3.86
So,
pH = 3.86 + log [0.10] / [0.13]
pH = 4.74 + log 0.769
pH = 4.74 - 0.11
pH = 4.63
