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Mass = 3 g
Volume = 3 mL
Density = mass / volume
D = 3 / 3
D = 1,0 g/mL
Answer:
Aromatic Hydrocarbons
Explanation:
Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.
At constant temperature, if the volume of the gas decreased to the given value, the pressure increases to 2.16atm.
<h3>What is Boyle's law?</h3>
Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.
Boyle's law is expressed as;
P₁V₁ = P₂V₂
Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.
Given the data in the question question;
- Initial volume of the gas V₁ = 682mL = 0.682L
- Initial pressure of the gas P₁ = 1.33atm
- Final volume of the gas V₂ = 0.419L
- Final pressure of the gas P₂ = ?
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = ( 1.33atm × 0.682L) / 0.419L
P₂ = 0.90706Latm / 0.419L
P₂ = 2.16atm
Therefore, at constant temperature, if the volume of the gas decreased to the given value, the pressure increases to 2.16atm.
Learn more about Boyle's law here: brainly.com/question/1437490
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Answer:
587.5 L (Option A)
Explanation:
The reaction is:
2 C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
We know that in STP conditions 1 mol of any gas is contained in 22.4L
Then, we can make a rule of three, to determine the moles of produced carbon dioxide.
22.4L is the volume for 1 mol
376 L will be the volume for (376 . 1) 22.4 = 16.78 moles at STP conditions.
Stoichiometry is 16:25.
16 moles of CO₂ are produce by the reaction of 25 moles of O₂
Then, 16.78 moles of CO₂ were produced by (16.78 . 25) /16 = 26.2 moles.
Now, the rule of three again.
1 mol of oxygen gas is contained at 22.4L, at STP conditions
26.2 moles might be contained at (26.2 . 22.4)/1 = 587.5 L
