A particular uranium alloy has a density of 18.75 g/cm3. Please answer the following questions below, providing the explanation to your answers. a. What volume is occupied by a critical mass of 49 kg of this alloy? b. The critical mass can be decreased to 16 kg if the alloy is surrounded by a layer of natural uranium (which acts as a neutron reflector). What is the volume of such smaller mass? Compare your answers to the approximate volumes of a baseball, a volleyball, and a basketball.

a) Critical mass of 49 kg of Uranium: 2600 cm³b) Critical mass of 16 kg Uranium: 850 cm³

Comparison:

Baseball: 221 cm³Basketball: 7,238 cm³Baseball < Critical mass of 16 kg of uranium < critical mass of 49 kg of uranium < basketball.

Explanation:

To find the volumes of the two samples of uranium alloy, you must use the formula of density:

         

a) Critical mass of 49 kg

First, notice that 18.75g/cm³ is the same that 18.75 kg/dm³ (to convert from grams to kg you must divide by 1,000, and to convert from cm³ to dm³ you must divide by 1,000; thus, dividing both numerator an denominator by the same number leaves the fraction unchanged).

Now you can compute:

   

You can convert to cm³ multiplying by 1,000cm³/dm³.

Thus, volume = 2,600 cm³.

b) Critical mass = 16 kg

     

Volume = 850cm³

Compare with approximate volumes of a baseball, a volleyball, and a basketball:

You can calculate those volumes by using the formula for the volume of a sphere:

       

For the baseball: radius ≈ 3.75cm

     

For the basketball radius ≈ 12.0 cm

     

Thus, the comparison of the volumes is:

Baseball < Crittical mass 16 kg of Uranium 16 kg < Critical mass 49 kg of Uranium  < Basketball

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts

C. definite shape, definite volume. that is a solid.