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Mariulka [41]
3 years ago
12

Which of the following electron configurations is incorrect? A) S [Ne]3s³3p⁴ B) Sn [Kr]5s²4d¹⁰5p² C) Rb [Kr]5s¹ D) V [Ar]4s²3d³

E) I [Kr]5s²4d¹⁰5p⁵
Chemistry
1 answer:
Verizon [17]3 years ago
4 0

Answer:

A S [Ne]3s³3p⁴ instead of [Ne]3s²3p⁴

Explanation:

Hello,

To find the incorrect electronics configuration, we have to refer back to our periodic table or simply writing the electronic configuration of each element down follow principles guiding it such as Aufbau principle and Hund's rule.

a) S = [Ne] 3s³ 3p⁴ instead of [Ne] 3s²3p⁴

Option A is wrong.

S orbital can only accommodate a maximum of 2 electrons and in this case, 3s orbital is carrying 3 electrons. This has violated the rule.

b) Sn = [Kr] 5s² 4d¹⁰ 5p²

Option B is correct

c) Rb = [Kr] 5s¹

Option c is correct

d) V = 4s² 3d³

Option D is correct

e) I = [Kr] 5s² 4d¹⁰ 5p⁵

From the above, we can see that the answer is option A = S

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What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
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The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

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HELP................!!!!
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