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12345 [234]
2 years ago
13

Score! You manage to find a bottle of bromothymol blue and a few extra beakers. You take one of the empty beakers and add some o

f the first unlabeled solution and some indicator. The color changes to yellow. You then add some solution from the other unlabeled flask into this beaker and see the color change to blue. What are the identities of each unlabeled solution
Chemistry
1 answer:
mamaluj [8]2 years ago
6 0

Answer:

You manage to find a bottle of bromothymol blue and a few extra beakers. You take one of the empty beakers and add some of the first unlabeled solution and some indicator.

The color changes to yellow.

You then add some solution from the other unlabeled flask into this beaker and see the color change to blue.

What are the identities of each unlabeled solution?

Explanation:

Bromothymol blue is a dye and it is used as an indicator.

It is used as a pH indicator.

In acids, it becomes yellow n in color.

In bases, it turns blue.

You take one of the empty beakers and add some of the first unlabeled solution and some indicator. The color changes to yellow.

That means the unlabeled solution is an acid.

You then add some solution from the other unlabeled flask into this beaker and see the color change to blue.

It is a basic solution.

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Which of the following water solutions would likely have the highest boiling point?
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3 years ago
If the molar mass of NaCl is 58.44 g/mol, what amount of NaCl (in mol) is contained in 250.0 mg of NaCl?
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2 years ago
The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and
Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

<em>...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).</em>

<em />

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

<em>Moles KSCN = Moles Ag⁺ in the filtrate:</em>

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

<em>Total moles Ag⁺:</em>

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

<em>Moles of Ag⁺ in the precipitate:</em>

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

<em>Moles AsO₄ = Moles As:</em>

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

<em>Moles As₂O₃:</em>

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

<em>Mass As₂O₃:</em>

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

<em />

7 0
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