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Mars2501 [29]
3 years ago
6

A mixture of benzoic acid and naphthalene was dissolved in tert-butyl methyl ether and the resulting solution was extracted with

aqueous sodium bicarbonate. The lower layer was separated and to it was added concentrated hydrochloric acid, upon which a white precipitate formed. What was the structure of the precipitate?
Chemistry
1 answer:
Llana [10]3 years ago
5 0

Answer:

C₆H₅COOH Benzoic Acid

Explanation:

Here we are using the extraction method to separate  benzoic acid from naphtalene in the  ether solution  by reacting it with the base sodium bicarbonate to produce the salt sodium benzoate  ( naphtalene remains in the organic layer )  .

The sodium benzoate  is completely soluble in the aqueous layer being a salt, and  reacts with hydrochloric acid ( again an acid base reaction ) which precipitates the benzoic acid  since the it is insoluble in water hence  separating it.

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A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what wil
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Answer:

Volume : 1.25 L

Explanation:

We are given here that the volume ( V_1 ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n_2 ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.

Volume ( V_1 ) = 1.50 L,

Initial moles ( n_1 ) = 3.00 mol,

Final Volume ( n_2 ) = 3.00 - 0.5 = 2.5 mol

Applying the combined gas law, we can calculate the final volume ( V_2 ).

P_1V_1 / n_1T_1 = P_2V_2 / n_2T_2 - we know that the pressure and temperature are constant, and therefore we can apply the following formula,

V_1 / n_1 = V_2 / n_2 - isolate V_2,

V_2 = V_1 n_2 / n_1 = 1.50 L * 2.5 mol / 3.00 mol = ( 1.5 * 2.5 / 3 ) L = 1.25 L

The volume of the balloon will be 1.25 L.

6 0
3 years ago
Which process is used to produce gases from solutions of salts dissolved in water or another liquid?
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Explanation:

Option D electrolysis is the correct answer

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Does the size of an object<br> affect its rate of fall?
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2 years ago
EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?
AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq).

The equilibrium constant for this reaction is called the self-ionization constant of water:

K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}].

Note that water isn't part of this constant.

The value of K_w at 25 °C is 10^{-14}. How to memorize this value?

  • The pH of pure water at 25 °C is 7.
  • [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}
  • However, [\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3} for pure water.
  • As a result, K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14} at 25 °C.

Back to this question. [\text{H}^{+}] is given. 25 °C implies that K_w = 10^{-14}. As a result,

\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}.

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3 years ago
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