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3 years ago

Of the following metals, which is the LEAST reactive? A. K B. Ca C. Na D. Mg

Chemistry

2 answers:

emmainna [20.7K]3 years ago

6 0

The Answer Should Be D.Mg

sladkih [1.3K]3 years ago

3 0

Answer: D. Mg

Explanation:

Chemical reactivity is defined as the tendency of an element to loose of gain electrons.

Metals are the elements which lose electrons and hence, their chemical reactivity will be the tendency to lose electrons.

Their chemical reactivity increases as we move top to bottom in a group because the valence shell come gets far away from the nucleus. Thus, the loss of electron from the valence shell becomes easier due to lesser attraction between nucleus and valence electron.

Their chemical reactivity decreases as we move from left to right in a period .As, the electrons get added up in the same shell, the electron in the outermost orbital gets near to the nucleus. And hence, the electron will be difficult to lose.Thus, the chemical reactivity of metals decreases.

For the given options:

Potassium(K) belongs to Group 1 and period 4 of the periodic table.

Calcium (Ca) belongs to Group 2 and period 4 of the periodic table.

Sodium (Na)belongs to Group 1 and Period 3 of the periodic table.

Magnesium (Mg) belongs to Group 2 and Period 3 of the periodic table.

Order of reactivity of metals follow:

$K>Ca>Na>Mg$

Thus the LEAST reactive metal is Magnesium.

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3 years ago

27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?

kati45 [8]

Answer:

8.69 is the pH at the equivalence point

Explanation:

Formic acid, HCHO₂, reacts with NaOH as follows:

HCHO₂ + NaOH → NaCHO₂ + H₂O.

At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:

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<em>Moles: </em>0.0278L * 0.797mol/L = 0.02216moles

To reach the equivalence point it is necessary to add:

0.02216mol * (1L / 0.928mol) = 0.0239L

Total volume in the equivalence point:

0.0278L + 0.0239L = 0.0517L

Concentration: 0.02216moles / 0.0517L = 0.429M

The equilibrium of NaCHO₂, CHO₂⁻, in water is:

CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)

Where Kb, 5.56x10⁻¹¹ is defined as:

5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]

In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:

5.56x10⁻¹¹ = [X] [X] / [0.429M]

2.383x10⁻¹¹ = X²

4.88x10⁻⁶ = X = [OH⁻]

As pOH = -log [OH⁻]

pOH = 5.31

And pH = 14 - pH

pH = 8.69 is the pH at the equivalence point

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4 years ago