A 20.5 lb cast iron boat anchor displaces 1.33 qt of water when submerged. What is the density of the cast iron, in g/cm3?

1L=1.057 qt
1kg=2.205lb
1lb=453.6g

20.5lb X 453.6g/ 1lb = 9298.9g

1.33qt X 1L/1.057qt = 1.26L

1.26L X 1000cm3/1L = 1260cm3

9298.9g/1260cm3 = 7.38g/cm3

**now why my teacher answer is 7.39g/cm3???

I think you rounded too soon. Just leave those numbers in the calculator.

20.5 x 453.6 = 9298.8 g (slight difference)
1.33 x 1000/1.057 = 1258.3 cc
9298.8/1248.3 = 7.39 g/cc.

thank you so much....your good, I wish you be my teacher I be passing my class with no problem....

Your calculation is correct, and the calculated density of the cast iron is 7.38 g/cm3. The discrepancy between your answer and your teacher's answer could be due to rounding errors in the conversion factors or in the final result. It is also possible that your teacher used slightly different values for the conversion factors or employed a different approach to the calculation. In any case, a difference of 0.01 g/cm3 is usually considered negligible in such calculations, and both answers can be considered accurate within the level of precision provided.

To calculate the density of the cast iron, you correctly converted the mass of the anchor from pounds (lb) to grams (g) using the conversion factor of 453.6 g/ 1 lb.

However, there may be a rounding error in your calculation. Let's double-check your work:

20.5 lb X 453.6 g/ 1 lb = 9298.8 g (rounded to one decimal place)

You correctly converted the volume of water displaced from quarts (qt) to liters (L) and then to cubic centimeters (cm3). Here's the step again:

1.33 qt X 1 L/ 1.057 qt = 1.26 L
1.26 L X 1000 cm3/ 1 L = 1260 cm3

Finally, to calculate the density, divide the mass (in grams) by the volume (in cubic centimeters):

9298.8 g / 1260 cm3 = 7.38 g/cm3 (rounded to two decimal places)

Based on your calculations, the density of the cast iron is 7.38 g/cm3. It seems that your teacher made a rounding error by rounding 7.38868 to 7.39. Both answers are acceptable depending on the desired level of precision.