2. D = M V First, enter the givens into the density table. 48,000 tons 1. Find the density of a substance whose mass is 48,000 tons and volume is 0.000600 pL? 48,000 tons0.000600 pL? 0.000600 pL

3. D = M V Next, enter the units from the mass and volume into the left side of the density table. 1. Find the density of a substance whose mass is 48,000 tons and volume is 0.000600 pL? 48,000 tons = ? pL = ? tons 0.000600 pL No conversion is necessary.

4. D Use the formula triangle to determine the correct formula. 1. Find the density of a substance whose mass is 48,000 tons and volume is 0.000600 pL? V M ÷ MVMV x D =

5. M V Substitute the values into the equation. 1. Find the density of a substance whose mass is 48,000 tons and volume is 0.000600 pL? 48,000 tons = ? pL = ? tons 0.000600 pL D = M/V D = (48,000 tons) / (0.000600 pL)

6. D = M V Divide and round to the LEAST number of significant digits. 1. Find the density of a substance whose mass is 48,000 tons and volume is 0.000600 pL? 48,000 tons = ? pL = ? tons 0.000600 pL D = M/V D = (48,000 tons) / (0.000600 pL) D = 8.0 x 10 7 tons/pL

7. D = M V First, enter the givens into the density table. 49.7 lbs 1 Tm 3 2. Find the volume of a substance whose density is 49.7 lbs/Tm 3 and mass is 204,900 ng? 49.7 lbs/Tm 3 204,900 ng ? 204,900 ng

8. D = M V Next, enter the units from the density into the right side of the density table. 49.7 lbs 1 Tm 3 2. Find the volume of a substance whose density is 49.7 lbs/Tm 3 and mass is 204,900 ng? 204,900 ng = ? Tm 3 = ? lbs

9. D = M V If a conversion is necessary, complete it now. 0.006921 lbs 1 Tm 3 2. Find the volume of a substance whose density is 49.7 lbs/Tm 3 and mass is 204,900 ng? 204,900 ng = ? Tm 3 = ? lbs

10. Start with what you are looking for!! lbs Since lbs is an English unit and the given, ng, is a metric unit, Ms. Mullins must provide an English to metric conversion. Always look back at the instructions to find the conversions. Use: 1 lb = 453.59 g g 1 453.59

11. Bring the denominator unit up to the next numerator. 1 lb 453.59 g g

12. Convert grams to nanograms. 1 lb 453.59 g g ng

13. Since g is further left (so it is larger) than ng, g gets the one and ng gets the ten. T _ _ G _ _ M _ _ k h da unit d c m _ _ μ _ _ n _ _ p 123864975 1 lb 453.59 g g ng 1 10 9

14. Bring the denominator unit up to the next numerator. 1 lb 453.59 g g ng 1 10 9 ng When you have reached the given unit, ng, then insert the given numerical value. 204,900

15. 1 lb 453.59 g g ng 1 10 9 ng 204,900 SOLVE!! = 4.517 x 10  7 lbs

16. D = M V Enter the answer from the dimensional analysis into the density table. 49.7 lbs 1 Tm 3 2. Find the volume of a substance whose density is 49.7 lbs/Tm 3 and mass is 204,900 ng? 204,900 ng = ? Tm 3 = 4.517 x 10  7 lbs

17. D Use the formula triangle to determine the correct formula. 2. Find the volume of a substance whose density is 49.7 lbs/Tm 3 and mass is 204,900 ng? V M ÷ MDMD x V =

18. D = M V Substitute the values into the equation. 49.7 lbs 1 Tm 3 2. Find the volume of a substance whose density is 49.7 lbs/Tm 3 and mass is 204,900 ng? 204,900 ng Tm 3 V = M/D V = (4.517 x 10  7 lbs) / (49.7 lbs/Tm 3 ) = 4.517 x 10  7 lbs

19. D = M V Divide and round to the LEAST number of significant digits. 49.7 lbs 1 Tm 3 2. Find the volume of a substance whose density is 49.7 lbs/Tm 3 and mass is 204,900 ng? 204,900 ng Tm 3 V = M/D V = (4.517 x 10  7 lbs) / (49.7 lbs/Tm 3 ) = 4.517 x 10  7 lbs V = 2.24 x 10  5 Tm 3

20. D = M V First, enter the givens into the density table. 0.006921 lbs 1 Tm 3 3. Find the mass of a substance whose density is 0.006921 lbs/mm 3 and volume is 7.0835 x 10 7 mi 3 ? 0.006921 lbs/Tm 3 7.0835 x 10 7 mi 3 ? 7.0835 x 10 7 mi 3

21. D = M V Next, enter the units from the density into the right side of the density table. 0.006921 lbs 1 Tm 3 3. Find the mass of a substance whose density is 0.006921 lbs/Tm 3 and volume is 7.0835 x 10 7 mi 3 ? 7.0835 x 10 7 mi 3 = ? Tm 3 = ? lbs

22. D = M V If a conversion is necessary, complete it now. 0.006921 lbs 1 Tm 3 3. Find the mass of a substance whose density is 0.006921 lbs/Tm 3 and volume is 7.0835 x 10 7 mi 3 ? 7.0835 x 10 7 mi 3 = ? Tm 3 = ? lbs

23. Start with what you are looking for!! Tm 3 Since Tm 3 is a metric unit and the given, mi 3, is an English unit, Ms. Mullins must provide an English to metric conversion. Always look back at the instructions to find the conversions. Use: 1 m = 1.0936 yd

24. Tm 3 However, the conversion is not a cubic conversion so.... 1 m = 1.0936 yd (1 m) 3 = (1.0936 yd) 3 1 m 3 = 1.3079 yd 3

25. Tm 3 In order to use the conversion, Tm 3 must be converted to m 3

26. Since Tm 3 is further left (so it is larger) than m 3, Tm 3 gets the one and m 3 gets the ten. Tm 3 m3m3 T _ _ G _ _ M _ _ k h da unit d c m _ _ μ _ _ n _ _ p 1 10

27. Either count by threes Tm 3 m3m3 T _ _ G _ _ M _ _ k h da unit d c m _ _ μ _ _ n _ _ p 1 10 369363024181233272115 36

28. Or count between units and multiply by three. Tm 3 m3m3 T _ _ G _ _ M _ _ k h da unit d c m _ _ μ _ _ n _ _ p 1 10 123121086411975 36 12 x 3 = 36

29. Bring the denominator unit up to the next numerator. 1 Tm 3 10 36 m 3 m 3

30. Now use the English to metric cubic conversion. 1 Tm 3 1 m = 1.0936 yd (1 m) 3 = (1.0936 yd) 3 1 m 3 = 1.3079 yd 3 10 36 m 3 m3m3 1 1.3079 yd 3

31. Bring the denominator unit up to the next numerator. 1 Tm 3 10 36 m 3 m3m3 1 1.3079 yd 3 yd 3

32. Remember that you are working towards the given which is mi 3. 1 Tm 3 10 36 m 3 m3m3 1 1.3079 yd 3 yd 3 There isn’t a conversion between yards and miles, but there is between feet and miles, so first develop a conversion between yd 3 and ft 3. 1 yd = 3 ft (1 yd) 3 = (3 ft) 3 1 yd 3 = 27 ft 3 ft 3 27 1

33. Bring the denominator unit up to the next numerator. 1 Tm 3 10 36 m 3 m3m3 1 1.3079 yd 3 1 yd 3 27 ft 3 ft 3

34. Remember that you are working towards the given which is mi 3. 1 Tm 3 10 36 m 3 m3m3 1 1.3079 yd 3 yd 3 There is a conversion between feet and miles, so first develop a conversion between ft 3 and mi 3. 5280 ft = 1 mi (5280 ft) 3 = (1 mi) 3 1.47 x 10 11 ft 3 = 1 mi 3 ft 3 27 1 ft 3 mi 3 1.47 x 10 11 1

35. Bring the denominator unit up to the next numerator. 1 Tm 3 1 m 3 1 yd 3 1.47 x 10 11 ft 3 10 36 m 3 1.3079 yd 3 27 ft 3 1 mi 3 mi 3

36. 1 Tm 3 1 m 3 1 yd 3 1.47 x 10 11 ft 3 10 36 m 3 1.3079 yd 3 27 ft 3 1 mi 3 mi 3 7.0835 x 10 7 When you have reached the given unit, mi 3, then insert the given numerical value.

37. SOLVE!! 1 Tm 3 1 m 3 1 yd 3 1.47 x 10 11 ft 3 10 36 m 3 1.3079 yd 3 27 ft 3 1 mi 3 mi 3 7.0835 x 10 7 = 2.9487 x 10  19 Tm 3

38. D = M V Enter the answer from the dimensional analysis into the density table. 0.006921 lbs 1 Tm 3 3. Find the mass of a substance whose density is 0.006921 lbs/Tm 3 and volume is 7.0835 x 10 7 mi 3 ? 7.0835 x 10 7 mi 3 = 2.9487 x 10  19 = ? lbs Tm 3

39. D Use the formula triangle to determine the correct formula. 3. Find the mass of a substance whose density is 0.006921 lbs/Tm 3 and volume is 7.0835 x 10 7 mi 3 ? V M ÷ M = D  V x

40. D = M V Substitute the values into the equation. 0.006921 lbs 1 Tm 3 3. Find the mass of a substance whose density is 0.006921 lbs/Tm 3 and volume is 7.0835 x 10 7 mi 3 ? 7.0835 x 10 7 mi 3 = 2.9487 x 10  19 = ? lbs Tm 3 M = D  V M = (0.006921 lbs/Tm 3 ) (2.9487 x 10  19 Tm 3 )

41. D = M V Multiply and round to the LEAST number of significant digits. 0.006921 lbs 1 Tm 3 3. Find the mass of a substance whose density is 0.006921 lbs/Tm 3 and volume is 7.0835 x 10 7 mi 3 ? 7.0835 x 10 7 mi 3 = 2.9487 x 10  19 = ? lbs Tm 3 M = D  V M = (0.006921 lbs/Tm 3 ) (2.9487 x 10  19 Tm 3 ) M = (2.041 x 10  21 lbs)

42. Practice Problems: 1. Find the mass of a substance whose density is 3.544 x 10  6 dg/yd 3 and volume is 94,000 Mm 3 ? 2. Find the density of a substance whose mass is 1.87 x 10  6 cg and volume is 234,567 dam 3 ? 3. Find the volume of a substance whose density is 0.00807400 Gg/hm 3 and mass is 204,900 tons? 4. Find the mass of a substance whose density is 0.0007770 lbs/nL and volume is 4.7222 x 10 4 yd 3 ? 5. Find the density of a substance whose mass is 4.8 x 10 28 kg and volume is 640,800 TL? 6. Find the volume of a substance whose density is 3.04 x 10  23  g/  m 3 and mass is 0.2030405 oz?