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Dmitriy789 [7]
3 years ago
12

Give the coordination number, the charge of the central metal ion, and select the correct name in each coordination compound: A.

Na3[CoCl6] coordination number charge of the central metal ion name B. [Ni(CO)4] coordination number charge of the central metal ion name C. [Ni(NH3)3(H2O)3] (NO3)2 coordination number charge of the central metal ion name
Chemistry
1 answer:
Komok [63]3 years ago
4 0

The correct question is

Give the coordination number, the charge of the central metal ion, and select the correct name in each coordination compound:

Na3[CoCl6]

[Ni(CO)4]

[Ni(NH3)3(H2O)2] (NO3)2

Answer:

See explanation for details

Explanation:

Na3[CoCl6]

Name: Sodium hexacholorocobalt III

Charge on the complex: -3

Central metal ion name : cobalt III

coordination number: 6

[Ni(CO)4]

Name: tetracarbonyl nickel (0)

Charge on the complex: 0

Central metal ion name : nickel (0)

coordination number: 4

[Ni(NH3)3(H2O)2] (NO3)2

Name: Diaquatriaminenickel II nitrate

Charge on the complex: +2

Central metal ion name : nickel (+2)

coordination number: 5

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Answer:

because

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How would you prepare 100.0 ml of.400 m CaCl2 from a stock solution of 2.00 M CaCl2?
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C₀=2 mol/l
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v₁=100.0 ml = 0.1 l

c₁v₁=c₀v₀

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Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra
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Answer:

molar solubility in water = 2.412 * 10^-4  mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

 

The equation for the solubilization reaction of Mg(OH)2 can be given as:

 

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

 Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²  

<u>Step 2:</u> Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

 X = <u>2.412 * 10^-4 mol/L = solubility in water</u>

<u>Step 3</u>: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+  = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X *(0.130)²  

5.61*10^-11 = X * (0.130)^2

X = <u>3.32*10^-9 = solubility in 0.130 M NaOH </u>

<u>Step 4:</u> Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

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