It’s baron I got it right on edg
Answer:
The reactant/reagent that would be most atom economical is EtI (Ethy Iodide) and KOH (potassium oxide) as base
This is because the iodo group are weak base hence they have a good leaving character (i.e they are unstable on their own ) which would increase the rate of reaction and the strong base KOH give the most atom economical
Explanation:
Answer:
The empirical formula is SF6 (option E)
Explanation:
Step 1: Data given
Mass of sulfur = 3.21 grams
Mass of fluorine = 11.4 grams
Molar mass sulfur = 32.065 g/mol
Molar mass fluorine = 19.00 g/mol
Step 2: Calculate moles
Moles = mass /molar mass
Moles sulfur = 3.21 grams / 32.065 g/mol
Moles sulfur = 0.100 moles
Moles fluorine = 11.4 grams / 19.00 g/mol
Moles fluorine = 0.600 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
S: 0.100 / 0.100 = 1
F : 0.600 / 0.100 = 6
The empirical formula is SF6 (option E)
