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sweet [91]
3 years ago
8

A 1.00 liter solution contains 0.43 M hydrofluoric acid and 0.56 M potassium fluoride. If 0.280 moles of potassium hydroxide are

added to this system, indicate whether the following statements are true or false. (Assume that the volume does not change upon the addition of potassium hydroxide.) A. The number of moles of HF will increase.B. The number of moles of F- will decrease. C. The equilibrium concentration of H3O will increase. D. The pH will decrease. E. The ratio of [HF] / [F-] will remain the same.
Chemistry
1 answer:
kolbaska11 [484]3 years ago
6 0

Answer:

Answers are in the explanation

Explanation:

Equlibrium of HF in H₂O is:

HF + H₂O ⇄ F⁻ + H₃O⁺

Now, the KOH reacts with HF, thus:

KOH + HF →  F⁻ + H₂O

<em>That means after reaction, concentration of HF decrease increasing F⁻ concentration.</em>

Now, seeing the equilibrium, as moles of HF decrease and F⁻ moles increase, the equilibrium will shift to the left decreasing H₃O⁺ concentration.

For the statements:

A. The number of moles of HF will increase. <em>FALSE</em>. HF react with KOH, thus, moles of HF decrease

B. The number of moles of F- will decrease. <em>FALSE</em>. The reaction produce  F⁻ increasing its moles.

C. The equilibrium concentration of H₃O⁺ will increase. <em>FALSE. </em>The equilibrium shift to the left decreasing concentration of H₃O⁺

D. The pH will decrease. <em>FALSE</em>. As the H₃O⁺ concentration decrease, pH will increase

E. The ratio of [HF] / [F-] will remain the same. <em>FALSE</em>. Because moles of HF are decreasing whereas F- moles are increasing changing, thus, ratio.

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Answer:

The order of reactivity towards electrophilic susbtitution is shown below:

a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene

b. p-cresol>p-xylene>toluene>benzene

c.Phenol>propylbenzene>benzene>benzoic acid

d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene

Explanation:

Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.

For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.

The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.

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3 years ago
What is the significance of molecular formula​
Brilliant_brown [7]

Answer:

The molecular formula of a compound represents molecular mass of the substance. For example, the molecular mass of water molecule (H2O) is 18u.

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6 0
3 years ago
In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].
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Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose +  H^+\rightarrow  fructose+ glucose

The rate law of the reaction is given as:

R=k[H^+][sucrose]

[H^+]=0.01M

[sucrose]= 1.0 M

R=k[0.01M][1.0 M]..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

R'=[0.01 M][2.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

R'=[0.01 M][0.5 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}

R'=2.5\times R

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when [H^+] is changed to 0.001 M = R'

R'=[0.0001 M][1.0 M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}

R'=0.01\times R

If concentration of  [H^+] is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and[H^+] both are changed to 0.1 M = R'

R'=[0.1M][0.1M]..[2]

[2] ÷ [1]

\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}

R'=1\times R

If concentration when [sucrose] and[H^+] both are changed to 0.1 M than rate will be increased by the factor of 1.

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Answer:

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Mass conc of H2O from the balanced equation = 2x18 = 36g

From the equation,

4g of H2 produced 36g of H2O

Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O

7 0
3 years ago
What percentage of radioactive substance remains after two half-lives
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Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

3 0
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