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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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For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be prod
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Answer:

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Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of CS_2 = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of H_2S = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

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For the first reaction:-

NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}

The mole ratio of the reactants = 1 : 1

0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride

So, <u>Moles of ammonium chloride formed = 0.5 moles</u>

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(b)

For the first reaction:-

CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}

The mole ratio of the reactants = 1 : 4

It means

0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.

But available moles of S = 0.5 moles

Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

4 moles of S produces 1 mole of CS_2

Thus,

0.5 moles of S produces \frac{1}{4}\times 0.5 mole of CS_2

<u>Mole of CS_2 = 0.125 mole</u>

Molar mass of CS_2 = 76.139 g/mol

<u>Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g</u>

4 moles of S produces 2 moles of H_2S

Thus,

0.5 moles of S produces \frac{2}{4}\times 0.5 mole of H_2S

<u>Mole of H_2S = 0.25 mole</u>

Molar mass of H_2S = 34.1 g/mol

<u>Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g</u>

<u></u>

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