Physics, asked by shaswataray, 10 months ago

Calculate the equivalent capacitor between A & B. Plz.. do it step by step. ​

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Answered by Madankumar808103
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A solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solution

Answered by Anonymous
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