If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1450 new snow flakes every minute, how much snow (in kilograms) accumulates on your lawn per hour? assume an average snow flake has a mass of 1.60 mg.

=58.464 kg/hr
width= 21.0 feet
length= 20.0 feet
snowflake mass=1.60 milligrams
1 foot^2=1450 snowflakes per minute
area of lawn=length*width
area=(21.0)(20.0)
area=420 feet^2
1450 snowflakes at 1.60 milligrams per snowflake:
(1.60)(1450)=2320 milligrams
2320 milligrams to kilograms:
=(2320 mg)(1 gram/1000 mg)(1 kilogram/1000 grams)
=(2320)(1/1000)(1/1000)
=(2.32)(1/1000)
=0.00232 kg per minute
0.00232 kg per minute to kg per hour:
=(0.00232 kg/1 minute)(60 minutes/1 hour)
=(0.00232)(60)
=0.1392 kg per hour per square foot
For the total lawn's accumulation in kilograms per hour, multiply by area: 
=(0.1392 (kg/hr) per 1 foot^2)(420 feet^2)
=58.464 kilograms of snow accumulate on the lawn per hour

No table found. the question asks for covalent bond based on electronegativies