Answer:
camphor sublimates salt is soluble in water while sand does not sublime and does not dissolve in water you first heat the mixture in a beaker covered with a watch glass camphor will then accumulate on the watch glass then you dissolve the remaining mixture of sand and salt salt will dissolve forming a salt solution then you filter using a filter paper and a beaker the residue on the filter paper is sand while the filtrate is salt solution you then heat the salt solution so that it can evaporate leaving salt particles thus you will have obtained salt sand and camphor
Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
Answer: Objects with like charge repel each other.
Answer:
The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
Explanation:
Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.
Molar heat of molar vaporization of Trichlorofluoromethane = 24.8 kJ/mol
Temperature at which Trichlorofluoromethan boils , T= 296.95 K
The molar entropy of the evaporation of Trichlorofluoromethan :
The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.
