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Untitled - Kelly Walsh High School

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10 CHEMISTRY FOR THE UTTERLY CONFUSED<br />

Careful!<br />

Clearly any of these procedures, and many more, will work. In this case, we will<br />

use the first:<br />

Density (Mass)<br />

(Volume)<br />

We also need to perform a volume conversion. As with the mass conversion,<br />

there are many different correct volume conversions. In this case, the conversion<br />

will relate, in some way, to the length unit of feet. Thus, we might use 12 in <br />

1 ft. If we incorporate this, we get:<br />

This will leave us with ft 2 ⋅ in. This is not very useful. We need to remember that<br />

ft 3 means ft × ft × ft. We can use the following to cancel the ft 3 :<br />

(2.00 ft 3 )a<br />

12 in<br />

1ft<br />

ba12 in<br />

1ft<br />

An easier way of writing this would be:<br />

(97.5 lb)<br />

<br />

(2.00 ft3 g<br />

a453.59<br />

) 1lb b<br />

(2.00 ft3 12 in<br />

)a<br />

1ft b<br />

in<br />

ba12 b . . . in3<br />

1ft<br />

(2.00 ft3 12 in<br />

)a<br />

1ft b<br />

3<br />

3456 in3 (unrounded)<br />

In this calculation you must remember that not only are the units cubed in the<br />

in/ft conversion, but also the numbers.<br />

We can then add our in/ft conversion to our density calculation:<br />

Density (Mass)<br />

(Volume)<br />

(97.5 lb)<br />

(2.00 ft 3 )<br />

g 1ft<br />

a453.59 ba<br />

1lb 12 in b<br />

3<br />

Now we have gotten to the units of g/in 3 . Since we do not have the units we are<br />

seeking (g/cm 3 ), we need at least one more step. We can finish the problem in<br />

one more step if we know the relationship between inches and centimeters. If

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