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.
= 24.3
The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).
Thus,
0.790 × 24 u = 18.96 u
0.100 × 25 u = 2.50 u
0.110 × 26 u = <u>2.86 u</u>
TOTAL = 24.3 u
∴ The relative atomic mass of X is 24.3.
Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.
Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:
Now we have to calculate the energy required.
where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:
Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
C. Rubbing the balloon against your hair
A wax is a type of lipid. Waxes are nonpolar, so they are not soluble in water.
