Answer:
The answer is below
Explanation:
The separation technique is used for separating immiscible liquids.
When separating, the stopper has to be removed when draining the lower layer so as to prevent a vacuum. If vacuum is allowed, the draining rate will reduce and stop.
The liquid should be mixed by shaking the funnel and then opening the stopcock so as the vent out gases.
When near interface between the layers, you should set your eye level so that you do not drain up to the second layer.
After completely draining the first layer, the second layer should be collected in a new flask.
After mixing the solutions in a separatory funnel, the stopper should be removed and the liquid should be mixed thoroughly and the layers allowed to separate. When you get close to the interface between the layers, get eye level with the funnel and slow the draining until the first layer is collected. Switch to a new flask to collect the second layer.
Answer: The value of 
for chloroform is 
when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Explanation:
Given: Moles of solute = 0.793 mol
Mass of solvent = 0.758
As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.
Now, the values of 
is calculated as follows.
where,
i = Van't Hoff factor = 1 (for chloroform)
m = molality
= molal boiling point elevation constant
Substitute the values into above formula as follows.
Thus, we can conclude that the value of for chloroform is when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
To make sure no cracks are in the crucible and also to remove any moisture present in the crucible by the process of heating.
Answer:
0.1988 J/g°C
Explanation:
-Qmetal = Qwater
Q = mc∆T
Where;
Q = amount of heat
m = mass of substance
c = specific heat of substance
∆T = change in temperature
Hence;
-{mc∆T} of metal = {mc∆T} of water
From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.
For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?
Note that, the final temperature of water and the metal = 24°C
-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)
-{34 × c × (-68°C)} = 459.8
-{34 × c × -68} = 459.8
-{-2312c} = 459.8
+2312c = 459.8
c = 459.8/2312
c = 0.1988
The specific heat capacity of the metal is 0.1988 J/g°C
