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Arada [10]
3 years ago
13

The following Lewis diagram represents the valence electron configuration of a main-group element. This element is in group 2A A

ccording to the octet rule, this element would be expected to form a(n) with a charge of cation anion If X is in period 4, the ion formed has the same electron configuration as the noble gas The symbol for the ion is________
Chemistry
1 answer:
Paraphin [41]3 years ago
4 0

<u>Answer:</u> The symbol of the ion formed is Ca^{2+}

<u>Explanation:</u>

An ion is formed when a neutral atom looses or gains electrons.

  • When an atom looses electrons, it results in the formation of positive ion known as cation.
  • When an atom gains electrons, it results in the formation of negative ion known as anion.

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

The element present in Group 2-A and in period 4 is Calcium (Ca)

Electronic configuration of Ca atom:  1s^22s^22p^63s^23p^64s^2

This atom will loose 2 electrons to attain stable electronic configuration similar to Argon element (noble gas)

The electronic configration of Ca^{2+}\text{ ion}=1s^22s^22p^63s^23p^6

Hence, the symbol of the ion formed is Ca^{2+}

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Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

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\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

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 Then the answer is number 4. This is because ionic compound conducts electricity when it is dissolved in water.

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