How many MOLECULES of boron tribromide are present in 7.65 grams of this compound? How many GRAMS of boron tribromide are present in 5.63×10^22 molecules of this compound?
Added by Tanya S.
Step 1
65 grams of boron tribromide. Step 2: Convert the number of molecules to grams for 5.63×10^22 molecules of boron tribromide. **Step 1:** Given mass of boron tribromide = 7.65 grams Molar mass of boron tribromide (BBr3) = 80.6 g/mol Avogadro's number = 6.022×10^23 Show more…
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