Answer:
Option C. Triple the number of moles
Explanation:
From the ideal gas equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of mole
R is the gas constant
T is the absolute temperature.
Making V the subject of the above equation, we have:
PV = nRT
Divide both side by P
V = nRT / P
Thus, we can say that the volume (V) is directly proportional to both the number of mole (n) and absolute temperature (T) and inversely proportional to the pressure (P). This implies that and increase in either the number of mole, the absolute temperature and a decrease in the presence will cause the volume to increase.
Thus, the correct option is option C triple the number of moles. This can further be seen as illustrated below:
Initial volume (V1) = 12 L
Initial mole (n1) = 0.5 mole
Final mole (n2) = triple the initial mole = 3 × 0.5 = 1.5 mole
Final volume (V2) =?
From:
V = nRT / P, keeping T and P constant, we have:
V1/n1 = V2/n2
12/0.5 = V2/1.5
24 = V2/1.5
Cross multiply
V2 = 24 × 1.5
V2 = 36 L.
Thus Option C gives the correct answer to the question.
To separate oil from water, add a solution of soluble ionic salt to the solution. Distillation, which involves boiling and condensing the water, is another option, as well as partially freezing the oil and water combination and discarding one-fourth to one-third of the unfrozen water.
Separation of components of crude oil
Answer:
0.84 moles of oxygen are required.
Explanation:
Given data:
Mass of CO₂ produced = 37.15 g
Number of moles of oxygen = ?
Solution:
Chemical equation:
C + O₂ → CO₂
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 37.15 g/ 44 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of oxygen and carbon dioxide.
CO₂ : O₂
1 : 1
0.84 : 0.84
0.84 moles of oxygen are required.
Data: molar mass 470 g/mol
Percent composition:
Hg = 85.0%
Cl = 15.0%
Solution:
1) Convert % to molar ratios
A. Base: 100 g
=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol
Cl = 15.0 g / 35.45 g/mol = 0.4231 mol
B. divide by the higher number and round to whole number
Hg = 0.4325 / 0.4231 = 1.00
Cl = 0.4231 / 0.4231 = 1.00
=> Empirical formula = Hg Cl
2) Find the mass of the empirical formula:
HgCl: 200.59 g/mol + 35.45 g/mol = 236.04
3) Determine how many times is the empirical mass contained in the molecular mass:
470 g/mol / 236.04 = 1.99 ≈ 2
=> Molecular formula = Hg2 Cl2.
Answers:
Empirical formula HgCl
Molecular Formula Hg2Cl2
