Answer:
Colloidal can not be separated through filtration.
Suspension can be separated through filtration.
Explanation:
Colloidal:
Colloidal consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm. While the pore size of filter paper is 2μm. That's why we can not separate the colloidal through the filtration. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.
Suspension:
The particle size in suspension is greater than 1000 nm. The particles in suspension can be separated through the filtration. These particles are large enough and can be seen through naked eye.
Answer:
a) both substances are insoluble in water
b) both substances are soluble in ligroin
c) both substances suffer combustion, octane produces more CO₂ than hexene.
d) both substances are less dense than waterl, with hexene having the lowest density.
e) only hexene would react with bromine
f) only hexene would react with permanganate
Explanation:
a) both substances are non-polar and water is polar
b) both substances are non-polar and lingroin is non-polar
c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O
C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O
d) water = 997 kg/m³
ocatne = 703 kg/m³
hexene = 673 kg/m³
e) bromine test is used to detect unsaturations
f) permanganate test is used to detect unsaturations
Density = m/v therefore v = m/d v = 25/0.85 v = 29.4117... cm^3 ( = 2.94 * 10^-5 m^3
Answer:
Explanation:
In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:
Now we can identify the variables:
If we plug all the values into the equation:
And we solve for 
:
I hope it helps!
