Answer:
Explanation:
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In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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Answer:
they can carry cold arctic air, producing snow and freezing weather.
Explanation:
-w- hope it helps
Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=
106.16×1000
17.12
=0.00016moles
Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO
2
=
44.01×1000
56.77
=0.0013
Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H
2
O==
18.02×1000
14.53
=0.0008
Moles ratios
\frac{0.0013}{0.0008}=1.625
0.0008
0.0013
=1.625
\frac{0.0008}{0.0008}=1
0.0008
0.0008
=1
Hence molecular fomula
The empirical formula is C 4H 5.
The molecular formula C8H10
Answer:
Explanation:
Moles of = 1 mole
Moles of = 1 mole
Volume of solution = 1 L
Initial concentration of = 1 M
Initial concentration of = 1 M
The given balanced equilibrium reaction is,
Initial conc. 1 M 0M 1 M
At eqm. conc. (1-2x) M (2x) M (1+x) M
The expression for equilibrium constant for this reaction will be,
The =
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
Concentration of at equilibrium= (2x) M =