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
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:
MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-
The initial concentrations are as follows:
[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles
Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles
Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles
EF= lowest number of moles over each of the elements
So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1
Therefore Emperical formula= C3H3O
Since there are no given items, I will give a general answer. Energy....or the lack of it. Examples: Heat, electricity, force (when an item is moving and it impacts something, it heats up...friction is an example of this), etc
