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mamaluj [8]
3 years ago
15

Is clo− a stronger or weaker base than hydroxylamine?

Chemistry
2 answers:
Levart [38]3 years ago
7 0


Clo− is a stronger base than hydroxylamine
<span>Clo−</span> (Hypochlorite ion) which is composed of chlorine and oxygen is a stronger base than hydroxylamine.The stronger acid will have the weaker conjugate base. Hydroxylamine, as the stronger acid, will have the weaker conjugate base and the hypochlorite ion will be the stronger base.<span />






likoan [24]3 years ago
5 0

\text{ClO}^- is \boxed{\text{stronger base}} than hydroxylamine.

Further Explanation:

Acids and bases are defined in accordance with different theories that are mentioned below.

Arrhenius theory:

It states that substances that release hydrogen ions or protons in aqueous solutions are known as acids while those releasing hydroxide ions in solutions are termed as bases.

Bronsted-Lowry theory :

In accordance with this theory, substances that are capable to donate protons to other species are acids while bases are entities that gain protons.

Lewis theory :

According to this concept, electron-deficient species are acids while electron-rich species act as bases.

In \text{ClO}^-, highly electronegative oxygen atom has negative charge on it so it is more likely to accept protons easily and therefore acts as strong base. But no such charge is present on any electronegtaive atom in hydroxylamine (\text{NH}_2\text{OH}). Therefore \text{ClO}^- is stronger base than hydroxylamine.

Learn more:

1. What is produced when base is dissolved in water? brainly.com/question/11656906

2. Reason for the acidic and basic nature of amino acid: brainly.com/question/5050077

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Acid, base and salts

Keywords: acid, base, electrons, protons, Arrhenius theory, Lewis theory, Bronsted-Lowry theory, hydrogen, hydronium ion, hydroxide ion, gain, donate, ClO^-, stronger base, hydroxylamine, oxygen, electronegative, negative charge, NH2OH.

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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2
nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔  2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2  [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ]  ' ²

Substituting the new concentration terms ,

K ' = 1/2  [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ]  ²

K ' = 1/4  [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ]  ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' =   [ NO ] ² [ Br₂ ] /  [ NoBr ]  ²

Hence ,

K' = K

5 0
3 years ago
What is the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml?
Firdavs [7]

Answer: The final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,  

M_1 = molarity of stock CuSO_4 solution = 2.5 M

V_1 = volume of stock CuSO_4solution = 5 ml  

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V_1 = volume of diluted CuSO_4 solution = 750 ml

Putting in the values we get:

2.5\times 5=M_2\times 750

M_2=0.017

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3 years ago
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Answer:

Hey mate.....

Explanation:

This is ur answer.....

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Hope it helps!

Brainliest pls....

Follow me! ;)

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Hope this helps!
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