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Korolek [52]
3 years ago
6

calcium and strontium are close to one another on thw oeriodic table. what do you know about their ionization energies ,their el

ectron configurations and their charasteristics based on their position in the periodic table?

Chemistry
1 answer:
Firdavs [7]3 years ago
4 0

Both Ca and Sr are in Group 2 of the Periodic Table, with Sr below Ca (see image)..

<em>Ionization energies </em>

They are both metals, so they have relatively <em>low ionization energies</em>.

Sr is below Ca in the Group, so Sr has the lower ionization energy.

<em>Electron configurations </em>

Their electron configurations both end in <em>ns²</em>.

For example, the electron configuration of Ca is [Kr]<em>4s²</em> and that of Sr is [Kr]<em>5s²</em>.

<em>Physical chsrscteristics: </em>

  • metallic lustre
  • malleable
  • ductile
  • good conductors of heat and electricity

<em>Chemical characteristics</em>

Both metals react with

• <em>Hydrogen</em> to form hydrides: M + H₂ ⟶ MH₂

• <em>Oxygen</em> to form oxides: 2M + O₂ ⟶ 2MO

• <em>Nitrogen</em> to form nitrides: 3M + N₂ ⟶ M₃N₂

• <em>Halogens</em> to form halides: M + X₂ ⟶ MX₂

• <em>Water</em> to displace hydrogen: M + 2H₂O ⟶ M(OH)₂ + H₂

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Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5​
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The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

K_{a}=\frac{[H^{+}][HB] }{[reactant]}

[HB] is the concentration of base.

8 * 10^{-5} =\frac{x^{2}  }{0.1}\\\\\\x^{2} = 8 * 10^{-5}*0.1

x^{2} = 0.08 * 10^{-4}\\ \\x = 0.283*10^{-2}

Then

pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54

So the pH of the solution is 2.54.

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What volume would 56.2 mL of gas at 820 mm of Hg occupy at 720 mm of Hg?
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Given: 56.2 mL of gas

To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg

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