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3 years ago

Give the coordination number of the central metal ion in the following complex ions.

Chemistry

1 answer:

melomori [17]3 years ago

8 0

Answer:

a). Coordination no. of $[Ni(en)_2(NH_3)_2]^2+$ = 6

b). Coordination no. of $[Fe(en)(ox)Cl_2]^2-$ = 6

Explanation:

Coordination number is defined as number of donar atoms bonded to the central atom of the complex ion.

a). Coordination no. of $[Ni(en)_2(NH_3)_2]^2+$ = 6

en or ethylenediamine is a bidentate ligand.

In bidentate ligand, two atoms directly bonded to the central atom.

NH3 is a unidentate ligand.

So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand

= $2\times 2 + 2 = 6$

b). Coordination no. of $[Fe(en)(ox)Cl_2]^2-$ = 6

Ethylenediamine (en) is a bidentate ligand.

oxalate ion (ox) is also a bidentate ligand.

Cl is a unidentate ligand

So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand

= $2\times 2 + 2 = 6$

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A 7.12 L cylinder contains 1.21 mol of gas A and 4.94 mol of gas B, at a temperature of 28.1 °C. Calculate the partial pressure

GenaCL600 [577]

Answer:

$P_A=4.20atm\\\\P_B=17.1atm$

Explanation:

Hello!

In this case, since the equation for the ideal gas is:

$PV=nRT$

For each gas, given the total volume, temperature (28.1+273.15=301.25K) and moles, we can easily compute the partial pressure as shown below:

$P_A=\frac{n_ART}{V} =\frac{1.21mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_A=4.20atm\\\\P_B=\frac{n_BRT}{V} =\frac{4.94mol*0.082\frac{atm*L}{mol*K}*301.25K}{7.12L} \\\\P_B=17.1atm$

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2 years ago

The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react

Kazeer [188]

Answer:

The percent yield of chloro-ethane in the reaction is 82.98%.

Explanation:

$C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl$

Moles of ethane = $\frac{300.0 g}{30 g/mol}=10 mol$

Moles of chlorine gases = $\frac{650.0 g}{71 .0 g/mol}=9.1549 mol$

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

$\frac{1}{1}\times 9.1549 mol=9.1549 mol$ of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

$\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%$

The percent yield of chloro-ethane in the reaction is 82.98%.

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2 years ago

In an anatomy class, a student has been learning about the bones of the body. While taking a test, he needs to identify the ster

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Answer:

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2 years ago

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Sodium chloride is produced from its elements

Kisachek [45]

Answer:

Mass of Sodium = 574.75 g

Mass of Chlorine = 886.25 g

Explanation:

The balance chemical equation for the synthesis of NaCl is,

2 Na + Cl₂ → 2 NaCl

Step 1: <u>Find out moles of each reactant required,</u>

According to balance chemical equation,

2 moles of NaCl is produced by = 2 moles of Na

So,

25 moles of NaCl will be produced by = X moles of Na

Solving for X,

X = 25 mol × 2 mol / 2 mol

X = 25 moles of Na

Similarly for Cl₂,

According to balance chemical equation,

2 moles of NaCl is produced by = 1 mole of Cl₂

So,

25 moles of NaCl will be produced by = X moles of Cl₂

Solving for X,

X = 25 mol × 1 mol / 2 mol

X = 12.5 moles of Cl₂

Step 2: <u>Convert each moles to mass as;</u>

Mass = Moles × Atomic Mass

For Na,

Mass = 25 mol × 22.99 g/mol

Mass = 574.75 g

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Mass = 12.5 mol × 70.90 g/mol

Mass = 886.25 g

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