Experiment predicted observation A student has two unopened cans containing carbonated water. Can A has been stored in the garage () and can B has been stored in the fridge (). The student opens one can at the time, both cans make a fizz. A) The fizz will be the same for both cans
B) There is not enough information to predict which can will make the louder fizz
C) Can A will make a louder and stronger fizz than can B.
D) Can B will make a louder and stronger fizz than can A.

Can A will make a louder and stronger fizz than can B.

Explanation:

Temperature has a direct effect on gas solubility. We know that carbonated water contains carbon dioxide dissolved in water. The extent of dissolution or solubility of this gas is dependent on the temperature of the system.

As the temperature of the system rises, the solubility of gas in solution decreases. It follows that can A, having been stored in a garage is definitely at a higher temperature than can B stored in the refrigerator.

Since solubility of gases decreases with increasing temperature, the carbon dioxide in can A will be less soluble than in can B. This will cause can A to make a louder and stronger fizz when opened than can B.

they both have a y in them

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml