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Tatiana [17]
3 years ago
15

Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:

Chemistry
1 answer:
Leya [2.2K]3 years ago
5 0

Answer :

(a) The symbol, group number, and period number of the element is, Cd, 12 and 5 respectively.

(b) The symbol, group number, and period number of the element is, Ni, 10 and 4 respectively.

Explanation :

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

Number of electrons in an atom are determined by the electronic configuration.

To identify the block of the element after you get its electronic configuration:

(i) If the element belongs to s-block.

Group number = Number of valence electrons (or outermost shell electrons).

(ii) If the element belongs to p- block.

Group number = No. of valence electrons + 10 .i.e., 10 + np electrons + ns electrons.

(iii) If the element belongs to d-block.

Group no. = no. of electrons in (n-1) d subshell + no. of electron/s in ns shell.

(iv) If the element belongs to f-block, Group no. = 3.

(a) [Kr] 5s²4d¹⁰

From the given electronic configuration, we conclude that it has 12 valence electrons and belongs to d-block. So, it belongs to group number 12 (2+10).

The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=5). So, the period number is, 5.

Thus, the element which is present in 5th period and 12th group number is, cadmium (Cd).

(b) [Ne] 4s²3d⁸

From the given electronic configuration, we conclude that it has 10 valence electrons and belongs to d-block. So, it belongs to group number 10 (2+8).

The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=4). So, the period number is, 4.

Thus, the element which is present in 4th period and 10th group number is, nickel (Ni).

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Explanation:

Step 1:

Data obtained from the question. This include the following:

van 't Hoff factor (i) = 1 (since the compound is non-electrolyte)

Mass of compound = 33.2g

Volume = 250mL

Osmotic pressure (Π) = 1.2 atm

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Step 2:

Determination of the molarity of the compound.

The molarity, M of the compound can be obtained as follow:

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1.2 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 1.2 / (0.0821 x 298)

M = 0.049mol/L

Step 3:

Determination of the number of mole of compound in the solution. This can be obtain as follow:

Molarity = 0.049mol/L

Volume = 250mL = 250/1000 = 0.25L

Mole of compound =..?

Molarity = mole /Volume

0.049 = mole / 0.25

Cross multiply

Mole = 0.049 x 0.25

Mole of compound = 0.01225 mole.

Step 4:

Determination of the molar mass of the compound. This is illustrated below:

Mole of the compound = 0.01225 mole.

Mass of the compound = 33.2g

Molar mass of the compound =.?

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The theoretical yield of silver chloride, AgCl is 5.2109 g

The balanced equation for the reaction is given below:

<h3>3AgNO₃(aq) + AICI₃(aq) —> Al(NO₃)₃ (aq) + 3AgCl (s) </h3>

Next, we shall determine the mass of aluminum chloride, AICI₃ that reacted and the mass of silver chloride, AgCl produced from the balanced equation. This is illustrated below:

Molar mass of AICI₃ = 133.34 g/mol

Mass of AICI₃ from the balanced equation = 1 × 133.34 = 133.34 g

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Mass of AgCl from the balanced equation = 3 × 143.32 = <em>429.96 g</em>

<h3>SUMMARY</h3>

From the balanced equation above,

133.34 g of AICI₃ reacted to produce 429.96 g of AgCl.

Finally, we shall determine the theoretical yield of AgCl by the reaction of 1.616 g of AICI₃ as follow:

From the balanced equation above,

133.34 g of AICI₃ reacted to produce 429.96 g of AgCl.

Therefore,

1.616 g of AICI₃ will react to produce = \frac{1.616 * 429.96}{133.34} = 5.2109 g of AgCl.

Thus, the theoretical yield of silver chloride, AgCl is 5.2109 g

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