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xenn [34]
4 years ago
12

How many ml of cranberry juice is required to prepare a 25 ml 12% cranberry solution using a 20% cranberry stock solution?

Chemistry
1 answer:
Zolol [24]4 years ago
7 0
To answer this question, you need to equalize the content of the cranberry for both solutions. The content could be count by multiplying the percentage with the volume of the cranberry juice. Then the calculation would be:

content1= content2
concentration1 * volume1= concentration2 * volume2
12%* 25 ml  = 20%* volume2
volume2= 25ml * 12%/ 20%= 15ml
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Answer:

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Explanation:

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In this reaction, both two reactants contain nitrogen with a different oxidation number and produce only one product which contains nitrogen with a unique oxidation state. So, nitrogen is oxidized and reduced in the same reaction.

Nitrogen Undergoes a change in oxidation state from 4+ in NO2 to 0 in N2. It is reduced because it gains electrons (decrease its oxidation state). NO2 is the oxidizing agent (electron acceptor).

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(ii) Zn(s) +AgNO3(aq) => Zn(NO3)2(aq) + Ag(s)

Ag changes oxidation state from 1+ to 0 in Ag(s).

Ag is reduced because it gains electrons and for this reason and AgNO3 is the oxidizing agent (electron acceptor)

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Balanced equation:

Zn(s) +2AgNO3(aq) => Zn(NO3)2(aq) + 2Ag(s)

 

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