Answer:
Option (d) is correct
N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺
Explanation:
Total electrons for all the species = 10
So these are <u>iso electronic</u> with each other.
We know
Ionic radii ∝ 
- Si⁴⁺ has 14 protons and 10 electrons
- Mg²⁺ has 12 protons and 10 electrons
- N³⁻ has 7 protons and 10 electrons
- F⁻ has 9 protons and 10 electrons
- Iso electronic species with greatest number of protons have small size and vice versa.
- So Si⁺⁴ have smallest size and N³⁻ have largest in size
We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:
Kw = Kb * Ka
where Kw is always equal to 1x10^-14
so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:
1x10^-14 = 1.4x10^-9 * Ka
solve for Ka
Ka = 7.14x10^-6
Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.
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Answer:
Density is a physical property that is determined by dividing the mass of a given amount of a substance by its volume.
Explanation:
Answer:
Nitrogen atom is small in size than phosphorus thus the bond between its nucleus and valence electrons are stronger than phosphorus.
Hope it helps...｡◕‿◕｡
Answer:
Electrons will flow from left to right through the wire.
Pb^2+ ions will be reduccd to Pb metal.
The concentration of Sn2+ ions in the left compartment will increase.
Explanation:
Looking at the relative electrode potentials of the two metals
Sn= -0.14
Pb=-0.13
Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.
