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Zina [86]
3 years ago
7

Which of these requires accurate coefficients in a reaction?

Chemistry
1 answer:
aksik [14]3 years ago
6 0

Answer:

A: molar ratio

Molar ratios state the proportions of reactants and products that are used and formed in a chemical reaction.

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The lithosphere, hydrosphere, and atmosphere are the abiotic parts of the planet.
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Valeric acid, HC5H9O2 (Ka = 1.5 ✕ 10−5), is used in the manufacture of magnesium valerate, a nerve-calming agent. What is the hy
Studentka2010 [4]

Answer:

[H^+]=0.00332M

Explanation:

Hello,

In this case, considering the dissociation of valeric acid as:

HC_5H_9O_2 \rightleftharpoons C_5H_9O_2 ^-+H^+

Its corresponding law of mass action is:

Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}

Now, by means of the change x due to dissociation, it becomes:

Ka=\frac{(x)(x)}{0.737-x}=1.5x10^{-5}

Solving for x we obtain:

x=0.00332M

Thus, since the concentration of hydronium equals x, the answer is:

[H^+]=x=0.00332M

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"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

8 0
3 years ago
Read 2 more answers
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