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Afina-wow [57]
3 years ago
11

A sample of aspirin weighing 5.945 g became contaminated with 2.134 g of sodium sulfate. The resulting

Chemistry
1 answer:
ira [324]3 years ago
7 0

Answer:

The correct answer to the following question will be "62.9 %".

Explanation:

The given values are:

The aspirin's initial amount = 5.945 g.

and is polluted containing 2,134 g of sodium sulfate.

After extraction we provided 3,739 g of pure aspirin.

Now,

Percentage \ of \ recovery \ aspirin =\frac{Aspirin \ amount \ isolated \ after \ extraction}{initial \ amount \ of \ aspirin}\times 100

On putting the values in the above formula, we get

⇒                                                    =\frac{3.739}{5.945}\times 100

⇒                                                    =62.9 \ percent

Note: percent = %

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The detector is capturing  3.3*10⁸ photons per second. So, in 1 hour:

E=5.93*10^{-23} \frac{J}{proton} *3.3*10^{8} \frac{proton}{s} *\frac{60}{1} \frac{s}{minute} *\frac{60}{1} \frac{minute}{hr}

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The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

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