3 years ago

A student was given a piece of metal with a mass of 85.0 g. She placed it

Chemistry

1 answer:

Anastasy [175]3 years ago

6 0

Answer:

22mL

Explanation:

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: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne

olchik [2.2K]

Answer:

The Net reaction is

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
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Explanation:

From the Question we are told that the buffers are

$H_2C_6H_5O_7^ -$ and $HC_6H_5O_7 ^{ 2-}$

When NaOH is added the Net ionic reaction would be

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
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If you have 222 g of radioisotope with a half-life of 5 days, how much isotope would remain after 15 days?

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If the half life is 5 days, you would have to divide 222 by 2 3 times.
so 222/2 =111g
111/2 = 55.5g
55.5/2 = 27.75g

after four half life the mass would be
27.75/2
13.875g

the percent would be
13.875/222
= 6.25%

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