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Answer:
0.45 moles
Explanation:
The computation of the number of moles left in the cylinder is shown below:
As we know that
we can say that
where,
n1 = 1.80 moles of gas
V2 = 12.0 L
And, the V1 = 48.0 L
Now placing these values to the above formula
So, the moles of gas in n2 left is
= 0.45 moles
We simply applied the above formulas so that the n2 moles of gas could arrive
Explanation:
It is given that volume of is 300 mL and molarity is 0.200 M.
Volume of NaCl is 200 mL and molarity is 0.050 M.
The chemical reaction will be as follows.
for
is given as
.
As, molarity is number of moles present in liter of solution.
Hence, moles of (aq) will be calculated as follows.
moles of (aq) =
= 0.06 mol
=
= 0.120 M
Mole of =
= 0.010 M
Now, Q =
=
=
As, Q < hence, there will be no formation of
precipitate.