What pressure will be necessary to compress 2 liters of a gas at 1 atm to a volume of 0.05 liters if the temperature remains con
1 answer:
<h3>Answer:</h3>
40 atm
<h3>Explanation:</h3>- According to Boyle's law, the pressure and the volume of a fixed mass of a gas are inversely proportional at constant absolute temperature.
- That is;
- At varying pressure and volume,
P1V1 =P2V2
In this case,
Initial volume, V1 = 2 L
Initial pressure, P1 = 1 atm
New volume, V2 = 0.05 L
We are required to calculate the new pressure,
Rearranging the formula;
P2 = P1V1 ÷ V2
= (1 atm × 2L) ÷ 0.05 L
= 40 atm
Therefore, the pressure required is 40 atm
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Answer:
Explanation:
First thing is we have assume all the percents are grams so we have
68.279g C, 6.2760g H, 3.7898g N, and 21.656g O
Now convert each gram to moles by dividing the the molar mass of each element
68.279g/12.01g= 5.685 moles of C
6.2760g/1.01g= 6.214 moles of H
3.7898g N/14.01g= 0.271 moles of N
21.656g O/ 16.00g= 1.354 moles of O
Now to find the lowest ratios divide all the moles by the smallest number of moles you found, in our case, the smallest moles is 0.271 moles of N so divide everything by that....
5.685 moles/0.271 moles ------> ~21 C
6.214 moles/0.271 moles --------> ~23 H
0.271 moles / 0.271 moles ---------> 1 N
1.354 moles/ 0.271 moles ----------> ~5 O
So the empirical formula is C21H23NO5