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Maslowich
3 years ago
12

What is the identity of the element which had the following electron configuration? 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​

Chemistry
1 answer:
kati45 [8]3 years ago
5 0

Answer:

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

Explanation:

Step 1: Data given

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​

has 25 electrons.

This element has 2 electrons on the first shell, 8 electrons on the second shell, 13 electrons on the third shell and 2 electrons on the outer shell (valence electrons).

This means this element is part of group VII.

The element with 25 electrons, we can find on the periodic table, with atomic number 25.

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

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1 year ago
Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.
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Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

  • pH = - log [H₃O⁺]

  • 4.64 = - log [H₃O⁺]

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b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

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3 years ago
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When Na_{2}CO_{3} is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of Na_{2}CO_{3} into CuCl_{2}(aq.) will result precipitation of insoluble CuCO_{3}

Reaction: Na_{2}CO_{3}(aq.)+CuCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CuCO_{3}(aq.)

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