To find the empirical formula you would first need to find the moles of each element:
58.8g/ 12.0g = 4.9 mol C
9.9g/ 1.0g = 9.9 mol H
31.4g/ 16.0g = 1.96 O
Then you divide by the smallest number of moles of each:
4.9/1.96 = 2.5
9.9/1.96 = 6
1.96/1.96 = 1
Since there is 2.5, you find the least number that makes each moles a whole number which is 2.
So the empirical formula is C5H12O2.
Answer: 0.07868 mol H₂O
Explanation:
1) Chemical equation:
Cu₂O +H₂ → 2Cu + H₂O
2) mole ratios:
1 mol Cu₂O : 1 mol H₂ : 2 mol Cu : 1 mol H₂O
3) Convert 10.00 g of Cu to grams, using the atomic mass:
Atomic mass of Cu: 63.546 g/mol
number of moles = mass in grams / atomic mass = 10.00g / 63.546 g/mol
number of moles = 0.1574 mol
4) Use proportions
2mol Cu 0.1574 mol Cu
--------------- = ---------------------
1 mol H₂O x
⇒ x = 0.1574 mol Cu × 1 mol H₂O / 2mol Cu = 0.07868 mol H₂O
That is the answer
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Answer:
The amount of Chlorodecane in the unknown is 0.105nmols
Explanation:
a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.
The area of the first peak corresponding to Chlorohexane is 32434 units.
The area of the second peak corresponding to chlorodecane is 2022 units.
Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:
1.69 nmols of Chlorohexane gives 32434 units
How much of chlorodecane gives 2022 units
By cross multiplication;
Moles of Chlorodecane = 2022*1.69/32434
=0.105nmols
Answer : The mole fraction and partial pressure of 
and 
gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.
Explanation : Given,
Moles of 
= 1.79 mole
Moles of 
= 1.20 mole
Moles of = 3.71 mole
Now we have to calculate the mole fraction of and gases.
and,
and,
Thus, the mole fraction of and gases are, 0.267, 0.179 and 0.554 respectively.
Now we have to calculate the partial pressure of and gases.
According to the Raoult's law,
where,
= partial pressure of gas
= total pressure of gas = 5.78 atm
= mole fraction of gas
and,
and,
Thus, the partial pressure of and gases are, 1.54, 1.03 and 3.20 atm respectively.
