3.2 g KClO3
Explanation:
1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)
= 0.0032 mol C12H22O11
0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)
= 0.026 mol KClO3
Therefore, the minimum amount of KClO3 needed is
0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)
= 3.2 g KClO3
Tim and Dan should expect to measure simple evaporation to happen at the end of their experiment.
Answer: 1.15 moles
Explanation:
1 mole = 6.02214076*10^23 molecules
The element Bromine is known as the chemical symbol Br2