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Word Problems using Sig. Figs.

Word Problems using Sig. Figs. Word Problems using Sig. Figs. Ex. (1) Find the average of the following lengths: 23.8 cm, 23.7 cm, 23.8 cm, 22.9 cm, 23.1 cm, 23.3 cm, and 23.0 cm . [ when averaging data, use the addition/subtraction rule . ]. 23.8 cm 23.7 cm 23.8 cm

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Word Problems using Sig. Figs.

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  1. Word Problems using Sig. Figs.

  2. Word Problems using Sig. Figs. Ex. (1) Find the average of the following lengths: 23.8cm, 23.7cm, 23.8cm, 22.9cm, 23.1cm, 23.3 cm, and 23.0 cm. [when averaging data, use the addition/subtraction rule.] 23.8 cm 23.7 cm 23.8 cm 22.9 cm 23.1 cm 23.3 cm + 23.0 cm true number 163.6 cm  7 = 23.37142857 cm 23.4 cm 163.6 cm

  3. Word Problems using Sig. Figs. Ex. (2) What is the volume of a solid block that is 2.7cm long, 1.6cm wide, and 6.3cm high? V = L  W  H V = 2.7 cm  1.6 cm  6.3 cm V = 27.216 cm3 V = 27 cm3

  4. Word Problems using Sig. Figs. Ex. (3) What is the density of the solid block in question #2 if its mass is 72.906 g? d = m V d = 72.906 g 27 cm3 d = 2.700222222 g/cm3 d = 2.7 g/cm3

  5. Word Problems using Sig. Figs. Ex. (4) What is the volume of a solid cylinder that has a radius of 1.46 cm and a height of 7.41 cm? (V = r2h, use 3.14 for ) V = r2h V = 3.14(1.46 cm)2(7.41 cm) V = 49.59678984 cm3 V = 49.6 cm3

  6. Word Problems using Sig. Figs. Ex. (5) What is the density of the solid cylinder in question #4 if its mass is 111.649 g? d = m V d = 111.649 g 49.6 cm3 d = 2.250987903 g/cm3 d = 2.25 g/cm3

  7. Word Problems using Sig. Figs. Ex. (6) If a beaker and a marble have a mass of 120.637 g and the beaker has a mass of 110.427 g, then what is the mass of the marble? m = 120.637 g – 110.427 g m = 10.21 g m = 10.210 g

  8. Word Problems using Sig. Figs. Ex. (7) What would be the mass, in grams, of a solid sample of gold if it were the same size and shape as a standard size basketball which has an official volume of 2250 cm3? d = m V 19.3g/cm3 = X 2250 cm3 1 X = 43425 g X = 43400 g

  9. Word Problems using Sig. Figs. Ex. (8) What would be the mass, in lbs, of the gold sample from question 7? [1.00 lb = 454 g] ____________ = ____________ X 43400 g Cross multiply and then divide 1.00 lb 454 g X = 95.81497797 lbs X = 95.6 lbs

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