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crimeas [40]
3 years ago
13

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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the answer is A hope this helps

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Kung ang English ng Araw ay Sun ano naman ang Araw-Araw?​
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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    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

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

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

8 0
3 years ago
Read 2 more answers
If you have 222 g of radioisotope with a half-life of 5 days, how much isotope would remain after 15 days?
Dima020 [189]
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%
7 0
3 years ago
What is the change in the enthalpy of 120 g/s acid heated in a double pipe heat exchanger from 25°C to 100°C, if the average hea
telo118 [61]

Answer:

ojopjd

Explanation:

3 0
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