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denis23 [38]
3 years ago
10

Which of the following is an example of qualitative observation?Select one of the options below as your answer:. A. Joseph is ob

serving the color difference in the reaction mixture of each of the test samples to see what types of proteins are present in them.. B. Joseph is measuring the color of the reaction mixture colorimetrically to see the concentration of proteins in the sample.. C. Joseph is observing the color of the reaction mixture to see whether proteins are present in the given solution..
Chemistry
2 answers:
masya89 [10]3 years ago
8 0
The probable answer here is letter C. 
Letter A can't be the answer since when we talk about color differences there is a "difference involve. So it is observing the magnitude of the change in color. 
Letter B can't be the answer since when we involve colorimetry, there is a corresponding values to these readings. This is actually a quantitative test. 
Letter C is the answer since, we are just observing if there is a certain color appears after the reaction takes place. 
vekshin13 years ago
5 0
<h2>Answer : Option C) Joseph is observing the color of the reaction mixture to see whether proteins are present in the given solution.</h2><h3>Explanation :</h3>

An example of qualitative observation is the one where one uses the five senses to identify the changes in the reaction.

Here, when Joseph is studying a reaction mixture he is trying to observe a color change which will confirm that there is proteins present in the reaction mixture or not If there is a color change observed then it will confirm the presence of proteins.

Usually qualitative observations are those which can be easily predicted by using five senses.


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Select the correct answer. When an atom in a reactant loses electrons, what happens to its oxidation number? A. Its oxidation nu
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C. Its oxidation number increases.

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Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.
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1.68% is ionized

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C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

<em>Where [] are concentrations in equilibrium</em>

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized  over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>
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