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:
72.6 mL
Explanation:
A quick way to solve this titration problem when you have a monoprotic acid is to use the Dilution equation, M1V1=M2V2.
.589(x)=.821(52.1)
X=72.6 mL
Answer:
i believe that the answer is c
Explanation:
I think its c because because its used as a bed rock layer
Chemical reaction: 4PBr₃(g) → P₄(g) + 6Br₂<span>(g).
</span>Pressure equilibrium constant (Kp) express the relationship between product pressures and reactant pressures. The partial pressures of gases are used to calculate pressure equilibrium constant.
Kp = (p(P₄) · p(Br₂)⁶) ÷ p(PBr₃)⁴.
p(P₄) - partial pressure of phosphorus.
p(Br₂) - partial pressure of bromine.
Answer:
Actually there is no answer but there is so much theories but people mostly says Bug Bang
