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Paraphin [41]
3 years ago
8

Which of the following statements is true about the strength of the intermolecular forces in CH4 and NH3?

Chemistry
1 answer:
inn [45]3 years ago
4 0

Answer:- C) CH_4 because the NH bond is more polar than CH bond.

Explanations:- methane is tetrahedral and it is symmetrical as all the bonds are same. Also, the electron negativity difference of C and H is too low and the molecule is almost non polar and makes the methane to have London dispersion forces.

Ammonia is trigonal pyramidal due to the presence of lone pair of electrons on central nitrogen atom. Nitrogen is highly electron negative atom so the N-H bond is more polar. Also, hydrogen bonding is possible in ammonia as the hydrogen is bonded to more electron negative nitrogen atom.

First option is not correct as the inter molecular forces are weaker in methane as compared to ammonia. Methane has only London dispersion forces where as ammonia has dipole-dipole as well as hydrogen bonding.

Second option is also not correct because the carbon has less partial negative charge as compared to N due to less electron negativity difference of C and H atoms.

Last choice is also not correct since the hydrogen bonding is not present in methane.

So, the only and only correct choice is C) CH_4 because the NH bond is more polar than CH bond.

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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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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

4 0
3 years ago
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