Answer:
CO.
Explanation:
Assuming the given percentages are by mass, we can solve this problem via imagining we have <em>100 g of the compound</em>, if that were the case we would have:
Now we <u>convert those masses into moles</u>, using the<em> elements' respective molar masses</em>:
- 42.9 g of C ÷ 12 g/mol = 3.57 mol C
- 57.1 g of O ÷ 16 g/mol = 3.58 mol O
As the number of C moles and O moles is roughly the same, the empirical formula for the compound is <em>CO</em>.
<span>Carbon dioxide is a pure compound, consisting of two kinds of atoms that are bonded together.</span>
Answer:
Iron (lll) oxide reacts with carbon monoxide according to the equation: Fe203(s) + 3 CO(g)-2Fe(s) + 3 CO2(g) A reaction mixture initially contains 22.55 g Fe203 and 14.78 g CO.
Explanation:
I hope this helps (:
T½=18.72days
therefore t¾=18.72+½ of 18.72
we have 18.72+9.36=28.08days
Explanation:
(a) Formula that shows relation between
and
is as follows.
Here,
= 1
Putting the given values into the above formula as follows.
= 
= 
= 0.01316
(b) As the given reaction equation is as follows.

As there is only one gas so
,
= 1.20
Therefore, pressure of
in the container is 1.20.
(c) Now, expression for
for the given reaction equation is as follows.
![K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCaO%5D%5BCO_%7B2%7D%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D)
=
= \frac{x^{2}}{(a - x)}[/tex]
where, a = initial conc. of 
=
= 0.023 M
0.0131 =
x = 0.017
Therefore, calculate the percentage of calcium carbonate remained as follows.
% of
remained =
= 75.46%
Thus, the percentage of calcium carbonate remained is 75.46%.