The answer is: supersaturated solution.
A supersaturated solution contains more of the dissolved substance than could be dissolved by the solvent under normal circumstances.
A way to dissolve more sugar into a solution is heating a solution.
The more heat is added to a system, the more soluble a substance (in this example sugar) becomes.
The solution will become supersaturated if this solution is suddenly cooled at a rate faster than the rate of precipitation.
Answer:
<u><em>When balloon filled with the helium gas is left in the car at night along with the decrease in temperature the size of the balloon will get smaller or it will look like little deflated</em></u>.The above observation can be explained on the basis of Charles law:The law states that 'under constant pressure ,the volume occupied by the gas is directly proportional to the temperature of the gas'. (At constant pressure)Temperature desecrate volume will also decreases and vice versa.So, with decrease in temperature at night the temperature of the gas in the balloon will also get decreased due to which the volume of the balloon will also get decreased.
Explanation:
The density told you the mass covered by one unit of volume of that substance.
Density defines how perfectly the molecules of a substance are packed in a unit of volume.
In the given problem the substances are expressed in g/ml which means that a unit of volume will be 1 mL.
Given ,
Density of mercury = 13.546 g/ml
It means 1mL of mercury will have a mass of 13.546 g, Similarly 1 mL of ethanol will have a mass of 0.789 g.
0.350 L * 10^3 ml/1 L* 13.546/1 mL = 4,741.1 g
The volume of ethanol will have an equal mass is
4741.1 g* 1 mL/0.789g = 6,008.9 mL
6,008.9 mL* 1L/10^3mL = 6.0081 L
Hence the answer is 6.0081 L
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Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration of high-pressure side is 
Explanation:
From the question we are told that
The thickness of the polyethylene is 
The temperature is 
The flux is 
The concentration on the low-pressure side is 
The initial diffusivity is 
The activation energy for diffusion is 
Generally the diffusivity of the oxygen at 600 K can be mathematically evaluated as
substituting values
Generally the flux is mathematically represented as
Where 
is the concentration of oxygen at the higher side
So
substituting values
