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pentagon [3]
2 years ago
8

The elements from this section of the periodic table all belong to the same

Chemistry
2 answers:
malfutka [58]2 years ago
8 0

Answer:

period

Explanation:

krok68 [10]2 years ago
7 0

Answer:

Option C = same period.

Explanation:

All these elements belongs to second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

Electronic configuration of lithium:

Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

Electronic configuration of boron:

B₅ = [He] 2s² 2p¹

Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell.

However their families, valance electrons and group are different. Boron have three valance electrons and belongs to group 3A. Carbon belongs to group 4A and have 4 valance electrons. Nitrogen belongs to group 5A and have five valance electrons. Oxygen belongs to group 6A and have six valance electrons. Fluorine belongs to group 7A and have seven valance electrons.

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Determine the molarity for each of the following solution solutions:
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Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of H_2SO_4 solution is, 0.00525 mole/L

(c)The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

(d)The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

(e)The molarity of Br_2 solution is, 0.0565 mole/L

(f)The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Solute is KCl.

\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of H_2SO_4, in 1.00 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is H_2SO_4

Molar mass of H_2SO_4 = 98 g/mole

\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L

The molarity of H_2SO_4 solution is, 0.00525 mole/L

<u>(c) 20.54 g of Al(NO_3)_3 in 1575 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is Al(NO_3)_3

Molar mass of Al(NO_3)_3 = 213 g/mole

\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L

The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

<u>(d) 2.76 kg of CuSO_4.5H_2O in 1.45 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is CuSO_4.5H_2O

Molar mass of CuSO_4.5H_2O = 250 g/mole

\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L

The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

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Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Solute is Br_2.

\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L

The molarity of Br_2 solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, C_2H_5NO_2, in 1.05 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

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Molar mass of C_2H_5NO_2 = 75 g/mole

\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L

The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

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