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MrRissso [65]
3 years ago
7

Which best explains why ionization energy tends to decrease from the top to the bottom of a group?

Chemistry
2 answers:
zheka24 [161]3 years ago
6 0

Answer:

Electrons get farther from the nucleus (C on edg)

Setler79 [48]3 years ago
5 0

Answer:

Electrons are far apart from the nucleus as we move down the group.

Explanation:

The ionization energy is the amount of energy which is necessary to remove an electron from an atom.

In an atom there exist a force of attraction at the center (nucleus). This is because of the positive charge which exists in the nucleus. This force of attraction is less felt as the distance between the electron and the proton increases. Hence the ionization energy increases as the number of shells increases for an atom. As we move down the group in the periodic table, the number of shells increases which implies a decrease in ionization energy.

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When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?
podryga [215]

Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.

c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

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Answer:

\boxed {\tt Number \ and \ unit}

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