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lesantik [10]
3 years ago
10

What will happen to the ability of a gas to dissolve in a liquid if the temperature increases

Chemistry
1 answer:
lord [1]3 years ago
4 0
It decreases i guess. just make sure,though i'm mostly correct.
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What is the valency of Aluminium and nitrate in Al(NO3)3 ?​
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Chlorofluorocarbons (CFCs) are no longer used as refrigerants because they destroy the ozone layer. Trichlorofluoromethane (CCl₃
Daniel [21]

Answer:

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

Explanation:

Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.

S=\frac{Q}{T(Kelvins)}

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Temperature at which Trichlorofluoromethan boils , T= 296.95 K

The molar entropy of the evaporation of Trichlorofluoromethan :

=\frac{24.8 kJ/mol}{296.95 K}=0.083516 kJ/mol K = 83.516 J/molK

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

8 0
2 years ago
A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

#SPJ1

7 0
1 year ago
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