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antiseptic1488 [7]
2 years ago
5

a gas occupies 3.5 l at standard presure. find the volume of the gas when the pressure is 1140 mm hg

Chemistry
1 answer:
Luda [366]2 years ago
8 0
According to Boyle's Law, " the pressure of gas is inversely related to applied pressure at constant temperature".

For the initial and final state of pressure and volume for a given gas at given temperature is as,

                                           P₁ V₁  =  P₂ V₂        ---------  (1)
Data Given;

                  P₁  =  760 mmHg (standard Pressure)

                  P₂  =  1140 mmHg

                  V₁  =  3.5 L

                  V₂  =  ?

Solving equation 1 for V₂,

                                           V₂  =  P₁ V₁ / P₂
Putting values,
                                           V₂  =  (760 mmHg × 3.5 L) ÷ 1140 mmHg

                                           V₂  =  2.33 L
Result:
           
Hence, with increase in pressure from 760 mmHg to 1140 mmHg the volume has decreased from 3.5 l to 2.33 L.
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Hello,

In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:

Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}

Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change x, we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

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