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Semenov [28]
3 years ago
13

How can you test the complete separation of camphor and sand​

Chemistry
2 answers:
Tcecarenko [31]3 years ago
8 0

Answer:

We know that camphor is a sublime substance and sand is not a sublime substance. Thus we can separate camphor from sand by heating the mixture slowly. The vapours of camphor should be then collected and cooled down. We would observe that solid camphor crystals will form.

Explanation:

katrin2010 [14]3 years ago
6 0

Answer: Use own words

What method can be used to separate iron filings and sand?

Iron is magnetic and the other two not, which means a magnet could be used to attract the iron filings out of the mixture, leaving the salt and sand. Salt is water soluble, while sand is not. This means the two can be mixed in water and stirred. The salt will dissolve and the sand will not.

other words

How will you separate Camphor common salt and sand?

Camphor can be separated using sublimation as it is a sublime,common salt and sand mixture can be separated using evaporation and filtration respectively. Principle : Throught sublimation we will get camphor, through decantation sand and then through evaporation salt.

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Penn State Civil and Environmental Engineering Department just built a super-duper coal-fired power plant in Altoona, PA. It is
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Answer:

<u>Part -A:</u>

Sulfur released is 1.338 \times 10^{4}pounds/hour.

<u>Part-B:</u>

SO_{2} released of per EPA norms is 4.865 /times 10^{-3}P/kw.

Explanation:

From the given,

The output = 1100 Mw

Efficiency = η = 0.4

Substitute the values in the following

Input = Output / η

= \frac{1100}{0.4}= 2750 Mw

Let's converts between joules to BTU.

1 BTU = 1056 J

<u>Part-A;</u>

Energy\,\, required = Power \times time

2750 \times 10^{6} \,\, J/s \times 3600s = 9.9 \times 10^{12}J

Energy \,\, required = \frac{9.9 \times 10^{22}}{1056}= 9.375 \times 10^{9} BTU

Mass\,of\, coal\,required = \frac{9.375 \times 10^{9}}{14,000}=6.69 \times 10^{5}pounds

But it has 2%  sulfur

The mass of sulfur released

6.69 \times 10^{5}pounds \,coal \times 0.02 = 1.338 \times 10^{4}pound

Therefore, released sulfur is 1.338 \times 10^{4}pounds/hour.

<u>Part -B;</u>

One pound of sulfur produce two pounds of sulfur dioxide

Initial amount of produced sulfur =

2 \times 1.338 \times 10^{4}pound = 2.676\times 10^{4}pound/hour

Assuming we added a 80% efficiency then,

Released sulfur dioxide = (1-0.8) \times SO_{2} \,produced =0.2 \times 2.676 \times 10^{4}Pounds/hr= 5.352 \times 10^{3}Pounds/hr

Energy produced in an hour = 400 mw \times 1 hr = 1.1 \times 10^{6}kwh

SO_{2}\,released \, as \, per EPA = \frac{5.352 \times 10^{3} \,pounds}{1.1 \times 10^{6}kwh}= 4.865 \times 10^{-3}p\kwh

Therefore, SO_{2} released of per EPA norms is 4.865 /times 10^{-3}P/kw.

8 0
3 years ago
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