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2 years ago

How many carbon atoms are there in one molecule of CH3FO?

Chemistry

1 answer:

Vlad1618 [11]2 years ago

8 0

Answer:

Brainliest pls

Explanation:

1 mole of a substance contains Avagadro’s number of particles,

i.e. 6.023*10^23

By unitary method,

5 moles of oxygen contains 5 times the Avagadro’s number of particles

i.e. 5* (6.023*10^23) = 3.0115*10^24 number of particles.

Now, the further answer depends on what particles the question concentrates on.

If number of atoms are asked , the above answer must be multiplied by 2, because oxygen is a diatomic gas and each atom contributes to be a particle.

therefore, 5 moles of oxygen has 6.023*10^24 atoms.

If number of molecules asked, the above answer is directly written...

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spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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4 years ago

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3 years ago

What is the percent by mass of oxygen in carbon dioxide

san4es73 [151]

C = 12
O2 = 16*2= 32
CO2 = (12)+(16*2) = 44

32/44*100 = 72.73%

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Answer:

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Explanation:

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Answer:

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