A mixture of CO2 and Kr weighs 35.0 g and exerts a pressure of 0.708 atm in its container. Since Kr is expensive, you wish to recover it from the mixture. After the CO2 is completely removed by absorption with NaOH(s), the pressure in the container is 0.250 atm. How many grams of CO2 were originally present? How many grams of Kr can you recover?

Explanation:

Since, it is given that carbon dioxide is completely removed by absorption with NaOH. And, pressure inside the container is 0.250 atm.

For Kr = 0.250 atm and pressure will be calculated as follows.

            = (0.708 - 0.250) atm

                       = 0.458 atm

Now, we will calculate the mole fraction as follows.

                       = 0.646

               Kr =

                    = 0.353

Now, we will convert into gram fraction as follows.

                          = 28.424

                   Kr =

                        = 29.57

Therefore, total mass is calculated as follows.

           Total mass = (28.424 + 29.57)

                              = 57.994

Hence, the percentage of and Kr are calculated as follows.

                     = 49%

               Kr =

                    = 51%

Hence, amount of and Kr present i mixture is as follows.  

          in mixture =

                             = 17.15 g

                  Kr =

                       = 17.85 g

Thus, we can conclude that 17.15 g of is originally present and 17.85 g of Kr is recovered.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.