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topjm [15]
2 years ago
13

The electron configuration can be also be represented by writing the symbol for the occupied subshell and adding a superscript t

o indicate the number of electrons in that subshell. For example, the electron configuration of the carbon atom having six electrons is written as
Chemistry
1 answer:
hichkok12 [17]2 years ago
4 0

Answer: The electron configuration of carbon atom is written as 1s²,2s², 2p².

Explanation:

The electronic configuration of an atom is defined as the arrangement of electrons into the shells or orbit of an atom. The constituents of an atom are proton, neutron and electron. the nucleus of an atom, where most of its mass are concentrated, consists of neutrons and protons fused together. Electrons occupy the shells surrounding the nucleus. The shells are lettered K, L, M, N and so on. Numerically, K shell is numbered 1, L is 2 and so on. These numbers also correspond with the increase in the energy level. All the electrons in K shell for instance belong to the first energy level and they have equal energy.

There is a limit to the number of electrons that can be found in a shell. This can be obtained by a formula 2n² where "n" is the energy level number of the shell.

K: 2n² = 2 × 1² = 2

L: 2n² = 2× 2² = 8

These shells are further subdivided into subshells. There are 4 subshells, s, p, d, and f. Each subshell can hold a different number of electrons.

This electron configuration of carbon can be written as 1s² 2s² 2p² where 1s, 2s, and 2p are the occupied subshells, and the superscript "2" is the number of electrons in each of these subshells.

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Groups of elements that are shiny but not lustrous, have a semi-conductive property, and are brittle are classified as?
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Answer:
A. Nonmetals
3 0
3 years ago
Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

  • If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

  • At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

  • At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0
3 years ago
What is the time it takes the Earth to make one revolution around the sun called?
Pavel [41]

Answer:

C

Explanation:

5 0
3 years ago
Read 2 more answers
2). H2S and SO2 as follows,(8pt)
Naddik [55]

Answer: 75.7 % yield

Explanation:

The balanced chemical equation is:

2H_2S+SO_2\rightarrow 3S+2H_2O  

According to stoichiometry :

68.2 g of H_2S will require = 64 g of SO_2

Thus 7.5 g of H_2S will require = \frac{64}{68.2}\times 7.5=7.0g of SO_2

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 68.2 g of give = 96 g of

Thus 7.5 g of H_2S give =\frac{96}{68.2}\times 7.5=10.5g  of S

% yield=\frac{\text{Actual yield}}{\text {Theoretical yield}}\times 100=\frac{7.95g}{10.5g}\times 100=75.7\%

Thus 75.7 % yield is there.

5 0
2 years ago
AP chem. Sublevels!!!!!!!!!!!
kumpel [21]

Answer:

i think its letter E. 2d

Explanation:

2d

4 0
2 years ago
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