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igomit [66]
2 years ago
6

Calculate the number of grams of ethanol nevessary to prepare 268g of a 9.76m solution of ethanol in water

Chemistry
1 answer:
stiv31 [10]2 years ago
7 0

The amount, in grams, of ethanol needed to prepare 268 grams of a 9.76 M solution of ethanol in water will be 120.50 grams

<h3>Solution preparation</h3>

In order to prepare 268 grams of solution, 268 mL of water would be needed because 1 mL of water weighs 1 gram.

Since we now know the volume of solution that we want to prepare, the amount of solute (ethanol) that would be required in order to make a solution of 9.76 M will be:

Mole required = 9.76 x 268/1000 = 2.62 moles

Mass of 2.62 moles ethanol = 2.62 x 46.07 = 120.50 grams

More on solution preparation can be found here: brainly.com/question/14667249

#SPJ3

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he decomposition of acetaldehyde, CH3CHO, was determined to be a second order reaction with a rate constant of 0.0771 M-1 s-1. I
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Answer:

The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M

Explanation:

Let the concentration of CH3CHO after selected reaction times be y

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The value of y must be positive and is obtained in terms of t using the quadratic formula

y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M

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