The empirical formula of the compound is calculated as follows
first calculate the mass of oxygen= 12-(4.09 +3.71)= 5.02g
then calculate the moles of each element, moles = mass/ molar mass
moles of K = 4.09g/39 g/mol(molar mass of K) = 0.105 moles
moles of Cl = 3.71g/35.5 g/mol(molar mass of Cl) = 0.105 moles
moles of O = 5.02g/ 16g/mol(molar mass of O) = 0.314 moles
then calculate e mole ratio by dividing each mole by the smallest number of moles ( 0.105 moles)
K=0.105/0.105= 1
Cl=0.105 /0.105=1
O= 0.314/0.105=3
therefore the empirical formula = KClO3
Answer:
Explanation:
You have an acid that is acidic or a base that is basic. When you mix the two, they form water (assuming those are bronsted-lowry acids and bases) which is neutral.
Answer:
92.344mL
Explanation:
acording to boyle's law that PV=constant then P1V1=P2V2
Answer:
900 J/mol
Explanation:
Data provided:
Enthalpy of the pure liquid at 75° C = 100 J/mol
Enthalpy of the pure vapor at 75° C = 1000 J/mol
Now,
the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.
Thus, mathematically,
The heat of vaporization at 75° C
= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C
on substituting the values, we get
The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol
or
The heat of vaporization at 75° C = 900 J/mol
