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andrew11 [14]
3 years ago
6

PLEASE HELP !

Chemistry
1 answer:
mario62 [17]3 years ago
4 0

Answer:

2L of water.

Explanation:

To know the volume of water to be added to the initial solution, first let us calculate the volume of the final solution. This is illustrated below:

Data obtained from the question:

Initial volume (V1) = 2L

Initial concentration (C1) = 6mol/L

Final concentration (C2) = 3mol/L

Final volume (V2) =?

Using the dilution formula, we can obtain the final volume of the stock as follow:

C1V1 =C2V2

6 x 2 = 3 x V2

Divide both side by 3

V2 = (6 x 2)/3

V2 = 4L.

The final volume of the solution is 4L.

To obtain the volume of water added, we shall determine the change in the volume of the solution. This is illustrated below:

Initial volume (V1) = 2L

Final volume (V2) = 4L

Change in volume = V2 – V1 = 4 – 2 = 2L.

Therefore, 2L of water must be added to the initial solution.

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What is an alkali metal with fewer than 10 protons in its nucleus?
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Answer:

            Lithium

Explanation:

                  Alkali metals are group of metals which are present in first group of periodic table. As we know atomic number is equal to number of protons contained by a particular element. Therefore, the alkali metals along with there number of protons are listed below;

Alkali Metal                                         Number of Protons

Lithium                                                               3

Sodium                                                             11

Potassium                                                         19

Rubidium                                                         37

Cesium                                                             55

Francium                                                          87

Hence, it is cleared from above table that Lithium is having fewer protons than 10.

6 0
2 years ago
If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
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Answer:

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