Don’t have a calculator on me but multiply 6.02x10 to the 23rd power by 4.5
Answer:
Explanation:
in a combustion of ethane 2 moles of ethane react with 7 moles of O2
now no of moles in 54 gram of O2=mass/ molar mass
moles =54/32=1.7 moles
if 7 moles of O2 required 2 moles of ethane then 1.7 mole required=?
7 moles of O2=2 moles of C2H6
1.7 moles of O2=1.7*2/7=0.5 moles of C2H6
0.5 moles of C2H6 contain how much grams=?
mass= moles*molar mass=0.5*30=15
Answer: but thx for the points
Explanation:
The Answer is A
Hope this helped ;)
Ignoring the n's, there would only be one unpaired electron.
