Apologia Chemistry: Yet More Density and Bowling for Density

BTW, five of the eight students have earned 100 on the 50 elements they need to memorize.  Yes, after taking the test five times, only one more kid learned them all today.  I share your frustrations teaching.  Chemistry is going sloowwwwwlllly.

Today, my husband, Rob, took over Chemistry while I taught some graphing calculator exercises and population pyramid histograms to Stat.  He spent the entire class redoing the density lab--but with calipers and introducing Bowling for Density.  Really.

After class, two of the kids, Lacie and James, spent two more hours with Bowling for Density.  The homework is to determine whether a 10 lb, 12 lb, or 14 lb bowling ball will float.  The density of water is 1.0 g/cm3.  If the bowling ball is less than 1.0 g/cm3, it will float.  If the ball's density is greater than 1.0 g/cm3, it will sink.  Sounds clear.  It isn't.

Bowling balls are 27 inches in circumference.  Their weight ranges from 10 lbs. to 16 lbs.  These are league specifications.  Density is mass/volume.  Since the density of water is 1.0 g/cm3, the mass must be in grams and the volume in cm3.  The volume of a sphere is here.  The first step is convert 27 inches into cm or 68.58 cm.

27 inches = 68.58 cm
 Circumference = 2πr
 68.58 = 2πr
 r = 10.9 cm
V = 4/3πr 3
 V = 4/3π(10.9)3
 V = 5446.8 cm3

Then one must convert lbs into grams:
 453.6 g/lb
10 lb x 453.6 = 4536

Density = mass/volume
4536 g /5446.8 cm3 = 0.83 g/cm3

0.83 g/cm3 is less than the density of water, 1.0 g/cm3.  Therefore, a 10 lb bowling bowl will float.

Here are samples of the work remitted.

The D scribbled at the top is my initial.  It means that I've entered the grade into Classroom.