Answer:
Carbon - 14
Oxygen - 16
Nitrogen - 15
Sulphur - 16
Explanation:
The question above is related to the "Periodic Table of Elements" which shows the proper arrangement of elements in a table. Every element falls on a<em> group/family</em> within the table. Each group has its own number, and the table has a total of<u> 18 groups</u><em> (from left to right). </em>They are classified according to <em>similarities in their characteristics</em>. For example, group 1 is composed of <em>alkali metals</em> while group 2 is composed of<em> alkali earth metals</em>.
Answer:
Explanation:
H₂SO₄ is a strong acid, which means that most of it ionizes in aqueous solution.
Since it is a diprotic acid (two hydrogen ions) its ionization occurs in two steps:
- H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
- HSO₄⁻ (aq) → H⁺(aq) + SO₄²⁻(aq)
Thus, almost all H₂SO₄ has ionized and its final concentration is almost nothing.
After the first ionization, the conentrations of H⁺(aq) and HSO₄⁻ are equal but by the second ionization more H⁺ ions are produced along with SO₄⁻.
You can show it as one step dissociation, assuming 100% dissociation (given this is a strong acid):
By the stequiometry you can build this table:
H₂SO₄ (aq) → 2H⁺(aq) + SO₄²⁻(aq)
Initial A 0 0
Change - x +2x +x
Equilibrium A - x 2x x
As explained, A - x is very low, and 2x is twice x. Thus,
The rank of the concentrations from highest to lowest is:
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
,
![p[CO_{2}] = K_{p}](https://tex.z-dn.net/?f=p%5BCO_%7B2%7D%5D%20%3D%20K_%7Bp%7D)
= 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%.
