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Pavlova-9 [17]
1 year ago
15

Your start-up company has invented a new water filter. If the filter can process 10 gallons of water per minute, how many gallon

s of water can it process in one 24-hour day?.
Chemistry
1 answer:
Marta_Voda [28]1 year ago
3 0

14,400 gallons of water can be filtered by the new water filter invented by the company.

It is given that the filter can process 10 gallons of water per minute.

So In an hour:

10 gallons per minute x 60 minutes = 600 gallons in 1 hour.

And in a day:

24 hours x 600 gallons per hour = 14,400 gallons in 24 hours.

Hence, 14,400 gallons of water can be filtered.

To know more about this gallons click here,

brainly.com/question/9560744

#SPJ4

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Answer: The answer is critical point

Explanation:

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What is the IUPAC name of the following compound?
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Answer:

(d) 3,7-dimethyl-4-nonyne.

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3 years ago
Which sentence explains why gases can change volume but liquids cannot?
choli [55]

Answer: Molecules of gas are usually far apart and can be compressed unlike molecules of liquids.

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Read 2 more answers
What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor
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Answer:

1.75272\ \text{m}^3

Explanation:

The breakdown reaction of ozone is as follows

CF_3Cl + UV \rightarrow CF_3 + Cl

Cl + O_3 \rightarrow ClO + O_2

O_3 + UV \rightarrow O_2 + O

ClO + O \rightarrow Cl + O_2

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of CF_3Cl = 15.5 g

M = Molar mass of CF_3Cl = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = 62.363\ \text{L mmHg/K mol}

Number of moles is given by

n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}

20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}

From ideal gas law we have

PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3

For 20 cycles of the reaction the volume of the ozone is 1.75272\ \text{m}^3.

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