Answer: 0.0022m3
Explanation:
Initial volume of mercury V1 = 0.002m3 (note that the unit of volume can be cm3, m3, dm3 or liters)
Initial temperature T1 = 20°C
Convert temperature in Celsius to Kelvin
( 20°C + 273°C = 293K)
Final temperature T2 = 50°C
( 50°C + 273°C = 323K)
Final volume V2 = ?
According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.
Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2
0.002/293 = V2/323
To get the value of V2, cross multiply
V2 = (323 x 0.002) / 293
V2 = 0.646 / 293
V2 = 0.0022 m3
Thus, the new volume of mercury will be 0.0022m3
The solution that will have the lowest freezing point is 5.0 SODIUM CHLORIDE.
Adding solute to solvents usually result in the depression of the freezing point. The higher the quantity of the solute that is added, the lower the freezing point of the solution.
The working equation for this is:
Tbp,soln - Tbp,water = i*Kb*m
where
Kb for water is 0.512 °C/molal
m is the molality (mol solute/kg solvent)
i is the van't hoff factor which represents the number of ions dissociated for strong electrolytes
Tbp,water is the boiling point of water which is 100°C
1. <span>1.50 moles of lioh (strong electrolyte) and 3.00 moles of koh (strong electrolyte) each in 1.0 kg of water
i = 2 for LiOH and 2 for KOH
Then,
</span>Tbp,soln - 100 = (2+2)(0.512)((1.5+3)/1 kg)
Tbp,soln = 109.22°C
<span>
2. </span><span>0.40 mole of al(no3)3 (strong electrolyte) and 0.40 mole of cscl (strong electrolyte) each in 1.0 kg of water
</span>
i = 4 for al(no3)3 and 2 for cscl
Then,
Tbp,soln - 100 = (4+2)(0.512)((0.4+0.4)/1 kg)
Tbp,soln = 102.46°C
<em>Thus, the first solution will have a higher boiling point.</em>
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 84.4 %
Percentage of H = 15.6 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 84.4 g
Mass of H = 15.6 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.03 moles.
For Carbon = 
For Hydrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H = 1 : 2
Hence, the empirical formula for the given compound is 
