Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
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Answer: The correct option is (c). The total pressure doubles.
Solution:
Initially, only 4 moles of oxygen gas were present in the flask.
(
) ( according to Dalton's law of partial pressure)
....(1)
= Total pressure when only oxygen gas was present.
Final total pressure when 4 moles of helium gas were added:
partial pressure of oxygen in the mixture :
Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.
= Total pressure of the mixture.
from (1)
On rearranging, we get:
The new total pressure will be twice of initial total pressure.
Answer:
C₃H₅O₂
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O
Explanation:
The reaction can be expressed as:
CₓHₓOₓ + nO₂ → CO₂ + H₂O
Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:
All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:
Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:
0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O
In order to determine the empirical formula, we calculate the moles of each component:
- mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
- mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
- mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O
Then we divide those values by the lowest one:
0.0236 mol C ÷ 0.0141 = 1.67
0.0354 mol H ÷ 0.0141 = 2.51
0.0141 mol O ÷ 0.0141 = 1
If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.
- The reaction is:
4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O