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andreev551 [17]
3 years ago
6

Acetone, CH3COCH3, is a nonelectrolyte; hypochlorous acid, HClO, is a weak electrolyte; and ammonium chloride, NH4Cl, is a stron

g electrolyte. Part A. What solute particles are present in an aqueous solution of CH3COCH3? Express your answer as a chemical expression. If there is more than one answer, separate them by a comma.
Chemistry
2 answers:
prisoha [69]3 years ago
5 0

Answer:

Part A: CH₃COCH₃

Explanation:

CH₃COCH₃ is a nonelectrolyte, which means it does not dissociate when it dissolves in water. Therefore, the only solute present will be the undissociated CH₃COCH₃.

Licemer1 [7]3 years ago
4 0
So the question ask on what solute are present in an aqueous solution on a certain element base on the data you have given. So base on that data i came up with a chemical expression of HCIO, H+ and CIO-. I hope you are satisfied with my answer 
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Read 2 more answers
What is Y? 222/86 Rn A/z + 4/2He
kondaur [170]

^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}

Y is Po-218.

  • A = 218
  • Z = 84.
<h3>Explanation </h3>

^{222}_{\phantom{2}86}\text{Rn} \to ^{A}_{Z}\text{Y}+ ^{4}_{2}\text{He}

Here's the symbol of a particle in a nuclear reaction. ^{A}_{Z}\text{Y}.

A stands for mass number. Z stands for atomic number. Both numbers shall conserve in a nuclear reaction.

  • The mass number on the left hand side is 222.
  • The two mass numbers on the right hand side add up to A + 4.
  • 222 = A + 4.
  • A = 218

So is the case for the atomic number. Try figure out the atomic number of Y using the same approach.

  • The atomic number on the left hand side is 86.
  • The two atomic number on the right hand side add up to __ + __.
  • 86 = __ + __.
  • Z = 84.

What element is Y? The atomic number of Y is 84. Refer to a periodic table. Element 84 corresponds to Po (polonium). Y is Polonium-218. The symbol of Y should be written as ^{218}_{\phantom{2}84}\text{Po}. Hence the equation:

^{222}_{\phantom{2}86}\text{Rn} \to ^{218}_{\phantom{2}84}\text{Po}+ ^{4}_{2}\text{He}.

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