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Gekata [30.6K]
3 years ago
6

Hydrogen sulfate reacts with water to produce sulfate ions and hydronium ions: hso4−+h2o⇌so42−+h3o+ identify the conjugate acid-

base pairs.
Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
8 0
The generic equation for a reaction between an acid and water is 

     HX_{(aq)} + H_{2}O_{(aq)} --> X_{(aq)}^{-} + H_{3}O_{(aq)}^+

When an acid "reacts" with water, water acts as the base that accepts the proton (H+) from the acid. The remaining ion that is formed after the acid has donated its proton is called the conjugate base (X^-), and the conjugate acid-base pair is HX - X^-. 

Hydrogen sulfate (HSO_{4}^-) is an ion from sulfuric acid. It is still an acid in itself and can "react" with water ((H_{2}O) to form the sulfate (SO_{4}^2-) and hydronium (H_{3}O^+)ions. 

HSO_{4(aq)}^{-} + H_{2}O_{(aq)} --> SO_{4(aq)}^{2-} + H_{3}O_{(aq)}^+

Based on the previous discussion, SO_{4(aq)}^2- is identified to be the conjugate of the acid HSO_{4(aq)}^-. Thus, the conjugate acid-base pair is HSO_{4(aq)}^{-} - SO_{4(aq)}^{2-}.
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An experiment with 55 co takes 47.5 hours. at the end of the experiment, 1.90 ng of 55-co remains. if the half-life is 18.0 hour
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Answer:

\boxed{\text{10.7 ng}}

Explanation:

Let A₀ = the original amount of ⁵⁵Co .

The amount remaining after one half-life is ½A₀.

After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.

After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.

The general formula for the amount remaining is:

A =A₀(½)ⁿ

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n = t/t_½

Data:

   A = 1.90 ng

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t_½ = 18.0 h

Calculation:

(a) Calculate n

n = 45/18.0 = 2.5

(b) Calculate A

1.90 = A₀ × (½)^2.5

1.90 = A₀ × 0.178

A₀ = 1.90/0.178 = 10.7 ng

The original mass of ⁵⁵Co was \boxed{\text{10.7 ng}}.

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