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:
Small-scale convection currents arise from uneven heating on a smaller scale. This kind of heating occurs along a coast and in the mountains. Small-scale convection currents cause local winds. Local winds blow over a much smaller area and change direction and speed over a shorter period of time than global winds.
Maybe that will help you answer the question.
B fossil energy
hope this helps!!
The mass in grams of butane at standard room temperature is 53.21 grams.
<h3>How can we determine the mass of an organic substance at room temperature?</h3>
The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:
PV = nRT
- Pressure = 1.00 atm
- Volume = 22.4 L
- Rate = 0.0821 atm*L/mol*K
- Temperature = 25° C = 298 k
1 × 22.4 L = n × (0.0821 atm*L/mol*K× 298 K)
n = 22.4/24.4658 moles
n = 0.91556 moles
Recall that:
- number of moles = mass(in grams)/molar mass
mass of butane = 0.91556 moles × 58.12 g/mole
mass of butane = 53.21 grams
Learn more about calculating the mass of an organic substance here:
brainly.com/question/14686462
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Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
