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xenn [34]
3 years ago
12

What is the volume of the air in a balloon that occupies 8.7 L at 50.0°C if the temperature is lowered to 25.0°C?

Chemistry
1 answer:
dangina [55]3 years ago
3 0

Answer:

8.03L

Explanation:

Given parameters:

Initial volume, V₁  = 8.7L

Initial temperature T₁  = 50°C , in Kelvin gives 50+273  = 323K

Final temperature T₂ = 25°C, in Kelvin gives 25 + 273  = 298K

Unknown:

Final volume V₂  = ?

Solution:

Since we are dealing with volume and temperature relationships under constant pressure, we simply apply the Charles's law.

Charles's law states that "the volume of a given mass of gas is directly proportional to the temperature provided the pressure remains constant".

Mathematically, we have;

              \frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }

where V andV T are volume and temperature, 1 and 2 are initial and final states.

Input the parameters and solve for V₂;

            \frac{8.7}{323}  =  \frac{V_{2} }{298}

          V₂  = 8.03L

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lutik1710 [3]
The first dissociation for H2X:
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initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


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Explanation:

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