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
What I’m seeing on quizlet says what you’re describing is a ball-and-stick model.
The correct answer is b. bimolecular
When the reaction equation is:
HF ↔ H+ + F-
and when the Ka expression
= concentration of products/concentration of reactions
so, Ka = [H+][F-]/[HF]
when we assume:
[H+] = [F-] = X
and [HF] = 0.35 - X
So, by substitution:
6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X
∴ X = 0.015 M
∴[H+] = X = 0.015
when PH = -㏒[H+]
∴PH = -㏒0.015
= 1.8
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
