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serg [7]
2 years ago
6

Write a net ionic equation to show that acetylsalicylic acid (aspirin), hc9h7o4, behaves as a brønsted-lowry acid in water.

Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
3 0
By definition, Bronsted-Lowry acid is a proton donor in the acid-base neutralization reaction. When a weak acid like acetylsalicylic acid is reacted with water, the water here acts as the Bronsted-Lowry base. This is possible because water has properties of amphoterism - can act as an acid or base. In this case, acetylsalicylic acid would have to donate its H+ atom to water, so that it would yield a hydronium ion, H₃O⁺. The complete net ionic reaction is shown in the picture.

So, in the reaction, the products yield are the acetylsalicylate ion and the hydronium ion.

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Which is a method that can be used to separate the components of a solution?
sashaice [31]

Answer:

Distillation

Explanation:

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4 0
3 years ago
A. 1720 kJ<br> B. 125.6 kJ<br> C. 3440 kJ<br> D. 4730 kJ
Feliz [49]

Answer:

Q = 3440Kj

Explanation:

Given data:

Mass of gold = 2kg

Latent heat of vaporization = 1720 Kj/Kg

Energy required to vaporize 2kg gold = ?

Solution:

Equation

Q= mLvap

It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg

by putting values,

Q= 2kg ×  1720 Kj/Kg

Q = 3440Kj

7 0
3 years ago
Convert 22.4 kg/L to kg/mL
liberstina [14]
1 kg/L -------------- 0.001 kg/mL
22.4 kg/L --------- ??

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7 0
3 years ago
What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

K_{a}=1.8\times 10^{-5}

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}

The expression for dissociation constant of acid is:

K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}

1.8\times 10^{-5}=\frac{x^2}{1.8-x}

1.8\left(1.8-x\right)=100000x^2

Solving for x, we get:

<u>x = 0.00568  M</u>

Percentage ionization = \frac{0.00568}{1.8}\times 100=0.32 \%

<u>Option B is correct.</u>

8 0
3 years ago
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