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An unknown element reacts with bromine to give bromide, MBr3. If 2.809g of the unknown element gives 27.765grams of MBr3, what...Asked by S-rth
An unknown element reacts with bromine to give bromide, MBr3. If 2.809g of the unknown element gives 27.765grams of MBr3, what is the element? ( give the symbol )
I don't understand the M in the MBr3...can someoen please help me with this question ASAP
thanks a lot
- i know this question was posted earlier but i did not understand something and bob did not reply to my question ( no prob , i understand you guys are busy but i really need help with this question, its worth a lot and im freaking out for the test tomm )
I don't understand the M in the MBr3...can someoen please help me with this question ASAP
thanks a lot
- i know this question was posted earlier but i did not understand something and bob did not reply to my question ( no prob , i understand you guys are busy but i really need help with this question, its worth a lot and im freaking out for the test tomm )
Answers
Answered by
Corey
The M stands for the unknown element; it binds to the Bromine. Bromine has a -1 charge, and there are three, it is highly likely that the element is Aluminum or Boron.
27.765 g MBr3-2.809 g M=24.956 g 3Br-
24.956 g Br* (1 mol Br/239.7 g)=
.104 mol Br
This is the same number as that of M, so there are .104 mol M.
2.809 g M/.104 mol M=26.98 g/mol M.
This is the molar mass of aluminum.
27.765 g MBr3-2.809 g M=24.956 g 3Br-
24.956 g Br* (1 mol Br/239.7 g)=
.104 mol Br
This is the same number as that of M, so there are .104 mol M.
2.809 g M/.104 mol M=26.98 g/mol M.
This is the molar mass of aluminum.
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