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Gemiola [76]
3 years ago
11

45 Three samples of the same solution are tested, each with a different indicator. All three indicators, bromthymol blue, bromcr

esol green, and thymol blue, appear blue if the pH of the solution is
(1) 4.7 (2) 6.0 (3) 7.8 (4) 9.9
Chemistry
2 answers:
zalisa [80]3 years ago
6 0
(4) 9.9 is the most sensible answer.

When an indicator appears blue, that means the solution is basic and the pH of the solution should be within the range of 8 and 14.

Hope this helps~


ANEK [815]3 years ago
3 0

<u>Answer:</u> The correct answer is Option 4.

<u>Explanation:</u>

Bromothymol blue, Bromocresol green and Thymol blue are the indicators which change their color according to the change in pH of the solution.

The pH range and color change of these indicators are:

  1. Bromothymol Blue: The pH range for this indicator is 6.0 to 7.5 and color change is from yellow to blue. It appears yellow below pH 6.0 and blue above pH 7.5
  2. Bromocresol green: The pH range for this indicator is 3.5 to 6.0 and color change is from yellow to blue. It appears yellow below pH 3.5 and blue above pH 6.0
  3. Thymol Blue: The pH range for this indicator is 8.0 to 9.6 and color change is from yellow to blue. It appears yellow below pH 8.0 and blue above pH 9.6

As, the highest pH of all the indicators is 9.6, so every indicator will appear blue above pH 9.6.

Hence, the correct answer is Option 4.

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if an exothermic reaction takes place in an insulated container at the end of the reaction the temperature of the content of the
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4 years ago
Determine the molarity and mole fraction of a 1.09 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of a
Juli2301 [7.4K]

Answer:

Molarity = 0.809 M

mole fraction = 0.047

Explanation:

The complete question is

Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution -

Solution for molarity:

1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.

1)  

Mass of 1.09 mole of acetone

= 1.09  mol x 58.0794 g/mol = 63.306 g

Density of acetone = 0.788 g/cm3  

Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3

For ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3  = 1.347 L

a) Molarity:

1.09 mol / 1.347 L = 0.809 M

Mole Fraction  

a) moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

b) moles of acetone:

1.09 / (1.09 + 21.71) = 0.047

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The process of centrifugation separates particles of size range between 0.1 μm to 5.0 μm efficiently.

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