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:
1) pure water
2) 0.75 m CaCl2
3) 1.0 m NaCl
4) 0.5 m KBr
5) 1.5 m glucose (C6H12O6)
Explanation:
Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.
The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.
The order of decreasing hoping point elevation is;
1) 0.75 m CaCl2
2) 1.0 m NaCl
3) 0.5 m KBr
4) 1.5 m glucose (C6H12O6)
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Answer:
(NH4)2S(aq) + Pb(NO3)2(aq) --> 2NH4NO3 (aq) + PbS (s)
