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
To find the third side you would use Pythagorean theorem which is a²+b²=c².
A and B being the 2 legs and C being the hypotenuse.
Example: ( sorry for the really odd not really a triangle, triangle...)
| \
6| \ ? 6²+9²=C²
|_______ 36+81=C²
9 117=C²
√117=10.8 (rounded)
if you need to get the leg you just fill in the numbers.
Answer:
sodium ,Na and magnesium,Mg
