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LuckyWell [14K]
3 years ago
5

Which statement is true about trends in metallic character?a) Metallic character increases as you go to the right across a row i

n the periodic table and as you go down a column.b) Metallic character decreases as you go to the right across a row in the periodic table and increases as you go down a column.c) Metallic character decreases as you go to the right across right across a row in the periodic table and decreases as you go down a columnd) Metallic characterdecreases as you go to the right across a row in the periodic table and increases as you go down a column
Chemistry
1 answer:
klasskru [66]3 years ago
6 0

Answer:

d) Metallic character decreases as you go to the right across a row in the periodic table and increases as you go down a column.

Explanation:

  • <em><u>Metallic character decreases as you move across a period in the periodic table from left to right.</u></em> This is because the attraction between valence electron and the nucleus increases, making it difficult for loss of electrons. It occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell.
  • On the other hand, <em><u>Metallic character increases as you move down an element group in the periodic table.</u></em> This is due to the increase in atomic radius as the number of energy levels increases.
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Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

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The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
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