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aev [14]
2 years ago
7

What is the coordination number for each of the following complexes or compounds? [Co(NH3)4(H2O)2]3+ [Cr(EDTA)]− [Pt(NH3)4]2+ Na

[Au(Cl)2]?
Chemistry
1 answer:
Burka [1]2 years ago
3 0

Answer:

[Co(NH₃)₄(H₂O)₂]³⁺: coordination number = 6.

[Cr(EDTA)]⁻: coordination number = 6.

[Pt(NH₃)₄]²⁺: coordination number = 4.

Na[Au(Cl)₂]: coordination number = 2.

Explanation:

  • [Co(NH₃)₄(H₂O)₂]³⁺:

In this complex, Co is bonded with 4 molecules of NH₃ (with 4 coordinate bonds, one bond for each molecule) and 2 molecules of H₂O (with 2 coordinate bonds, one bond for each molecule) forming the complex with 6 coordinate bonds.

∴ coordination number = 6.

  • [Cr(EDTA)]⁻:

In this complex, Cr is bonded with 1 molecules of EDTA (with 6 coordinate bonds, 4 O atoms and 2 N atoms in EDTA molecule).

∴ coordination number = 6.

  • [Pt(NH₃)₄]²⁺:

In this complex, Pt is bonded with 4 molecules of NH₃ (with 4 coordinate bonds, one bond for each molecule).

coordination number = 4.

  • Na[Au(Cl)₂]:

In this complex, Au is bonded with 2 atoms of Cl (with 2 coordinate bonds, one bond for each atom).

coordination number = 2.

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The density of Ca will be between that of Mg and Sr

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Ammonia and oxygen react to form nitrogen and water.
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A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

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