Zn = 28.15%
Cl = 30.53%
O = 41.32%
<h3>Further explanation</h3>
Given
Zn(CIO3)2 compound
Required
The % composition
Solution
Ar Zn = 65.38
Ar Cl = 35,453
Ar O = 15,999
MW Zn(CIO3)2 = 232.3
Zn = 65,38/232.3 x 100% = 28.15%
Cl = (2 x 35.453) / 232.3 x 100% = 30.53%
O = (6 x 15.999) / 232.3 x 100% = 41.32%
Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.
Explanation:
The reaction equation is as follows.

Initial : 0.3 0.4 0
Change: -x -x x
Equilbm: (0.3 - x) (0.4 - x) x
We know that, relation between standard free energy and equilibrium constant is as follows.

Putting the given values into the above formula as follows.


x = 0.1417
Hence, at equilibrium
= 0.1583 M
= 0.2583 M
Hi,
The statement is true, as the volume of a sample depends on its size.
I hope this helps. If I was not clear enough or if you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.
Answer:
a. Change of state
Explanation:
Because you will see that the state has changed