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Doss [256]
3 years ago
5

Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes formula hydroch

loric acid hcl sodium hydroxide naoh formic acid hcooh methyl amine ch3nh2 potassium chloride kcl ethanol c2h5oh sucrose c12h22o11
Chemistry
1 answer:
matrenka [14]3 years ago
3 0

Compounds which on dissolving in water gets completely dissociates into its ions are known as strong electrolytes whereas compounds which on dissolving in water gets partially dissociates into its ions are known as weak electrolytes.


Substances which gives solution on dissolving in water and do not dissociates into ions also does not conduct electric current are known as nonelectrolyte.

  • Hydrochloric acid, HCl

On adding HCl (strong acid) in water, it will completely dissociates into ions (H^{+} and Cl^{-}) and thus, it is a strong electrolyte.

  • Sodium hydroxide, NaOH

On adding NaOH (strong base) in water, it will completely dissociates into ions (Na^{+} and  OH^{-}) and thus, it is a strong electrolyte.

  • Formic acid, HCOOH

On adding HCOOH (weak acid) in water, it will partially dissociates into ions (H^{+} and  HCOO^{-}) and thus, it is a weak electrolyte.

  • Methyl amine, CH_3NH_2

On adding CH_3NH_2 (weak base) in water, it will partially dissociates into ions (CH_3NH_3^{+} and  OH^{-}) and thus, it is a weak electrolyte.

  • Potassium chloride, KCl

On adding KCl in water, it will completely dissociates into ions (K^{+} and  Cl^{-}) and thus, it is a strong electrolyte.

  • Ethanol, C_2H_5OH

On adding C_2H_5OH in water, it will not dissociates into ions  and thus, it is a nonelectrolyte.

  • Sucrose, C_{12}H_{22}O_{11}

On adding C_{12}H_{22}O_{11} in water, it will not dissociates into ions  and thus, it is a nonelectrolyte.

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If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theor
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Answer:

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27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

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To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

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Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.

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Answer:

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Explanation:

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