1.35 mole of K₂SO₄ contain 2.7 moles of cation.
Cations are positive ions. They are the ions of metallic elements.
To obtain the answer to the question, we'll begin by calculating the number of mole of cations in 1 mole of K₂SO₄.
K₂SO₄ (aq) —> 2K⁺ (aq) + SO₄²¯ (aq)
<h3>Cation => K⁺</h3>
From the balanced equation above,
1 mole of K₂SO₄ contains 2 moles of K⁺ (i.e cation).
Finally, we shall determine the number of mole of cation in 1.35 mole of K₂SO₄. This can be obtained as follow:
1 mole of K₂SO₄ contains 2 moles of K⁺ (i.e cation).
Therefore, 1.35 mole of K₂SO₄ will contain = 1.35 × 2 = 2.7 moles of K⁺ (i.e cation).
Hence, we can conclude that 1.35 mole of K₂SO₄ contain 2.7 moles of cation
Learn more: brainly.com/question/24707125
Answer : The concentration of 
at equilibrium is 0 M.
Solution : Given,
Concentration of = 0.0200 M
Concentration of 
= 1.00 M
The given equilibrium reaction is,
![Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3C_2O_4%5E%7B2-%7D%28aq%29%5Crightleftharpoons%20%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%28aq%29)
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of 
will be,
![K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%5D%7D%7B%5BC_2O_4%5E%7B2-%7D%5D%5E3%5BFe%5E%7B3%2B%7D%5D%7D)
By solving the term, we get:
Concentration of at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of at equilibrium is 0 M.
They are all <u>non-metal elements</u>.
They have a mass for the particles
There are no totally elastic collisions
There are intermolecular forces
