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ANEK [815]
4 years ago
8

What would be the most likely charge of an ion formed from an atom with the electron configuration: 1s22s22p63s2 (2 points) +1 -

1 +2 -2
Chemistry
2 answers:
Shalnov [3]4 years ago
4 0

Answer : The correct option is, (+2)

Explanation :

The given electronic configuration is, 1s^22s^22p^63s^2

In the given electronic configuration of an atom, '3s' shell is the outermost shell in which two electrons are present. For the stable configuration, these two electrons will remove easily from the outer shell and atom will carry (+2).

Total number of electrons in the given configuration= 12 electrons

And magnesium is the element which have 12 electrons. Due to the stable electronic configuration, magnesium atom changes their electronic configuration by losing two electrons and Mg atom changes to Mg^{2+} ion.

Hence, the most likely charge of an ion formed from an atom with the given electron configuration would be, (+2)

Lady bird [3.3K]4 years ago
3 0
+2 D. because this atom is magnesium and mag. has a charge of +2
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Answer:

14.53ML

Explanation:

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3 years ago
How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution
Lelechka [254]

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

  • The mole ratio of the acid, HCl (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
  • Volume of acid, HCl (Va) = 36 mL
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MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

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brainly.com/question/14356286

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2 years ago
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Step 2: Calculate the pH

We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.

pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6

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