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Troyanec [42]
2 years ago
14

10 points

Chemistry
1 answer:
artcher [175]2 years ago
7 0
C. arteries, veins, capillaries
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Which is a characteristic of a strong base
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3 0
3 years ago
The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0
2 years ago
Upon dissolving in , undergoes a disproportionation reaction according to the following unbalanced equation: This disproportiona
pychu [463]

Answer:

The question in the narrative seems not to be complete because the unbalanced equation was not given and the values of the second(s) and  chemical dissolved was also not given.

Kindly find the complete question below and if you feel the question is still correct, the solution provided is still implies but without the values inserted.

Correct Question:

Upon dissolving InCl(s) in HCl , In(aq)  undergoes a disproportionation reaction according to the following unbalanced equation:

In ⁺ (aq) → In(s) → In³⁺ (aq)

This disproportionation follows first-order kinetics with a half-life of 667s. What is the concentration of In⁺ (aq) after 1.25 h if the initial solution of In⁺ (aq) was prepared by dissolving 2.38 g InCl(s) in dilute HCl to make 5.00 x 10² mL of solution? What mass of In(s) is formed after 1.25h?

Solution / Explanation:

Given half life of In⁺ at 66.7 s,

We recall the formula used in claculating the rate of constant for the first oreder reactin as :

K = 0.693 / t₁÷2,

Noting that:

t₁÷2 = half life

and K= rate constant

Therefore, if we replace the value of  t₁÷2  in the formular above,

We have,

K = 0.693 / 667s

K = 0.00 /s

Now, if we recall the mass of InCl(s) as 2.38g,

Volume of dilute HCl = 500 mL,

and the molar mass of  InCl(s) - 150.271 g/mol,

The number of moles is then calculated using the formular:

Number of moles: = Given Mass/Molar Mass

Now replacing the given values of given mass and the molar mass in the above formular,

= 2.38g / 150.271g/mol

= 0.0158 mol

Volume of diluted 500 mL.

Recalling also that we need to convert from mL into Liters

Therefore,  1mL = 10⁻³L

Therefore,

500mL = (500 X 10⁻³)L

0.5 L

Now, the molarrity of In⁺ (aq) is calculated using

morality of In⁺ (aq) = moles of In⁺ (aq)/volume of solution

= 0.0158/0.5L

=0.0316M (This is the initial concentration of In⁺ (aq))

The time of the reaction is 1.25h

There is 3600s in one hour

1.25h = 1.25 x 3600

= 4500s

7 0
3 years ago
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