Stirring this is because the three elements are factors affecting dissolving of a solvent. Eg temprature affects in hotness or coldness, Particle size affects whether it is big or small while quantity of soluble affects by the amount
Answer:
The ration of molar concentration is "2.5".
Explanation:
The given values are:
Average density of salt water,
= 
Net pressure,
= 
Increase in pressure,
= 
Now,
The under water pressure will be:
= 
= 
= 
= 
hence,
The ratio will be:
= 
or,
= 
= 
= 
<span>Pass the mixture through filter paper. The large particles in the suspension will filter out. to tell the difference between a solution and a colloid, shine a beam of light through the mixture, if it reflects then it is a colloid, if it doesn't then it is a solution</span>
Answer:
The answers are in the explanation.
Explanation:
The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:
Increasing temperature of ice from -10°C - 0°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g
Q = 2.06J/g°C*10°C*10g
Q = 206J
Change from solid to liquid:
The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:
Q = 333.55J/g*10g
Q = 3335.5J
Increasing temperature of liquid water from 0°C - 100°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g
Q = 4.18J/g°C*100°C*10g
Q = 4180J
Change from liquid to gas:
The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:
Q = 2260J/g*10g
Q = 22600J
Increasing temperature of gas water from 100°C - 120°C:
Q = S*ΔT*m
Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g
Q = 1.87J/g°C*20°C*10g
Q = 374J
Total Energy:
206J + 3335.5 J + 4180J + 22600J + 374J =
30695.5J =
30.7kJ
Answer:
B
Explanation:
When water is at the surface of the ground, it evaporates and goes back to the clouds for which causes it to rain therefore returning water back.