Answer:
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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:
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(7.3 x 10^29 atoms) / (24 atoms/molecule) / (6.022 x 10^23 molecules/mol) =
5.1 x 10^4 mol C6H12O6
Ionic bonding is formed in Ionic compounds due to electrostatic force between the oppositely charged ions.
In covalent bonds electrons are shared between the atoms. In case of ionic bond the bond is stronger as there is complete transfer of electrons from one ion to the other.
Since the ionic bonds are more difficult to break than the covalent bonds, ionic compounds have a higher melting point than covalent compounds.