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
<span>C7H8
First, determine the number of relative moles of each element we have and the molar masses of the products.
atomic mass of carbon = 12.0107
atomic mass of hydrogen = 1.00794
atomic mass of oxygen = 15.999
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
We have 5.27 mg of CO2, so
5.27 / 44.0087 = 0.119749 milli moles of CO2
And we have 1.23 mg of H2O, so
1.23 / 18.01488 = 0.068277 milli moles of H2O
Since there's 1 carbon atom per CO2 molecule, we have
0.119749 milli moles of carbon.
Since there's 2 hydrogen atoms per H2O molecules, we have
2 * 0.068277 = 0.136554 milli moles of hydrogen atoms.
Now we need to find a simple integer ratio that's close to
0.119749 / 0.136554 = 0.876937
Looking at all fractions n/m where n ranges from 1 to 10 and m ranges from 1 to 10, I find a closest match at 7/8 = 0.875 with an error of only 0.001937, the next closest match has an error over 6 times larger. So let's go with the 7/8 ratio.
The numerator in the ratio was for carbon atoms, and the denominator was for hydrogen. So the empirical formula for toluene is C7H8.</span>
Answer:
A disproportionation reaction is when a multiatomic species whose pertinent element has a specific oxidation state gets oxidized and reduced in two separate half-reactions, yielding two other products containing the same pertinent element. A convenient example is Mn2O3 becoming Mn2+ and MnO2 .
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