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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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You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.220 M sodium benzoate
ddd [48]

Answer:

34.5 mL

Explanation:

Given, we ned to prepared buffer solution with pH = 4.00 , volume = 100.0 mL .

[C6H5COOH] = 0.100 M , [C6H5COO-] = 0.120 M , pKa = 4.20

First we need to calculate the ratio of the conjugate base and acid

We know, Henderson Hasselbalch equation

pH = pKa + log [C6H5COO-] /[C6H5COOH]

4.00 = 4.20 + log [C6H5COO-] /[C6H5COOH]

log [C6H5COO-] /[C6H5COOH] = 4.00-4.20

= - 0.20

Antilog from both side

[C6H5COO-] /[C6H5COOH] = 0.631

Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume

Sum of volume = 100 mL = 0.100 L

volume of benzoic acid = x

volume of benzoate = 0.10 -x ,

so Volume of acid + volume of conjugate base = 0.100 L

[C6H5COO-] /[C6H5COOH] = 0.631

[C6H5COO-] = 0.631 * [C6H5COOH]

0.120 (0.1-x) = 0.631 *0.100x

So, x = 0.065

So, volume of benzoic acid = x = 0.0655 L

= 65.5 mL

So, volume of sodium benzoate = 0.1 -x

= 0.1-0.0655

= 0.0345 L

= 34.5 mL

So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.

5 0
3 years ago
Read 2 more answers
A sample of an ionic compound that is often used as a dough conditioner is analyzed and found to contain 4.628 g of potassium, 9
Roman55 [17]

Answer:

The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

Explanation:

Mass of potassium = 4.628 g

Moles of potassium = \frac{4.628 g}{39 g/mol}=0.1187 mol

Mass of bromine = 9.457 g

Moles of bromine = \frac{9.457 g}{80 g/mol}=0.1182 mol

Mass of oxygen = 5.681 g

Moles of oxygen = \frac{5.681 g}{16 g/mol}0.3551

For empirical formula of the compound, divide the least number of moles from all the moles of elements present in the compound:

Potassium :

\frac{0.1187 mol}{0.1182 mol}=1.0

Bromine;

\frac{0.1182 mol}{0.1182 mol}=1.0

Oxygen ;

\frac{0.3551 mol}{0.1182 mol}=3.0

The empirical formula of the compound is = KBrO_3

The name of the compound is potassium bromate.

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Answer:

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Answer:

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