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3 years ago

Which orbital is the first to fill with electrons in a given principal energy level?

Chemistry

2 answers:

frez [133]3 years ago

8 0

The s orbital is first to fill

zepelin [54]3 years ago

4 0

Answer: s orbitals

Explanation: s orbital is the first to fill with electrons in a given principal energy level.

It has the most penetrating power out of all the 4 orbitals.

There are 4 orbitals in which the electrons are filled regularly.These are s(sharp), p(principle), d(diffused) and f(fundamental).

S orbitals are near the nucleus . The electrons are filled as per Aufbau Principle.

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deff fn [24]

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3 years ago

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3 years ago

What can be categorized as a use of plasma in technology

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3 years ago

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zhuklara [117]

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3 years ago

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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3 years ago