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djverab [1.8K]
3 years ago
8

Sodium is more reactive than magnesium. Give reason ​

Chemistry
2 answers:
scoray [572]3 years ago
4 0

Answer:

The alkaline-earth metals tend to lose two electrons to form M 2+ ions (Be2+, Mg2+, Ca2+, and so on). These metals are less reactive than the neighboring alkali metal. Magnesium is less active than sodium; calcium is less active than potassium; and so on. These metals become more active as we go down the column.

Morgarella [4.7K]3 years ago
3 0

Answer: According to the reactivity of metals Sodium is the second metal while Magneesium is the fourth metal. The metal in the top of the reactivity serise table are more reactive.

<em>(The series obtained by the arrangement of metals in the decending oder of their  reactivity is the REACTIVITY SERIES TABLE)</em>

Explanation:

K

<u><em>Na</em></u>

Ca

<u><em>Mg</em></u>

Al

Zn

Fe

Sn

Pb

H

Cu

Hg

Ag

Pt

Au

Therefore Sodium is more reactive than Magnesium

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The electronegativity is 2.1 for h and 1.9 for pb. based on these electronegativities pbh4 would be expected to
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Based on this:
PbH4 would be expected to <span>have polar covalent bonds with a partial negative charges on the H atoms. </span>
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ASAP: in group 13, the only metalloid, boron, is at the very top of the group. Explain why the rest of the group has metallic be
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3 0
3 years ago
Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base
lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = 6.20*10^{-5}

So:

pKa = - log\:(Ka)

pKa = - log\:(6.20*10^{-5})

pKa = 5 - log\:6.20

pKa = 5 - 0.79

\boxed{pKa = 4.21}

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

pH = pKa + log\:\dfrac{[salt]}{[acid]}

pH = 4.21 + log\:\dfrac{0.130}{0.235}

pH = 4.21 + log\:0.55

pH = 4.21 + (-0.26)

pH = 4.21 - 0.26

\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

8 0
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