Asked by Susie
5.00 mol of solid A was placed in a sealed 1.00-L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.10 M, where it remained constant. Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?
A(s)--> B(g)+C(g)
A(s)--> B(g)+C(g)
Answers
Answered by
guru
just multiply the concentration of
Bx2=1.10Mx2L=2.2moles
and then subtract the moles of A from calculated moles of B
5.00-2.2=2.8moles
Bx2=1.10Mx2L=2.2moles
and then subtract the moles of A from calculated moles of B
5.00-2.2=2.8moles
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