Please show workThank you! a. 8.24 m to mm(1 m = 1000 mm)b. 2.67 qt to L(1 qt = 0.947 L)c. 6.1 ft to cm(1 in = 2.54 cm)d. 6.475 cm2 to m2(100 cm = 1 m)e. 101 lb to kg(1 kg = 2.205 lb)%3Df. 101 lb to g(1 kg = 1000 g)g. 65 mph to km/hr(1 km = 0.6214 mi)%3D(1000 g = 2.205 lb)(1 ft = 30.48 cm)h. a density of 0.791g/cm3 to lb/ft3%3Di. a pressure of 14.7 lb/in2 to kg/m2(use a combination of the above)

This question is based on unit conversion. Thus,

a. 8.24 m to mm (1 m = 1000 mm) %3! b. 2.67 qt to L (1 qt = 0.947 L) c. 6.1 ft to cm (1 in = 2.54 cm) d. 6.475 cm2 to m2 (100 cm = 1 m) e. 101 lb to kg (1 kg = 2.205 lb) %3D f. 101 lb to g (1 kg = 1000 g) %3D g. 65 mph to km/hr (1 km = 0.6214 mi) h. a density of 0.791g/cm3 to Ib/ft3 (1000 g = 2.205 Ib) (1 ft = 30.48 cm) i. a pressure of 14.7 lb/in2 to kg/m² (use a combination of the above)

a. 8.24 m to mm (1 m = 1000 mm) %3! b. 2.67 qt to L (1 qt = 0.947 L) c. 6.1 ft to cm (1 in = 2.54 cm) d. 6.475 cm2 to m2 (100 cm = 1 m) e. 101 lb to kg (1 kg = 2.205 lb) %3D f. 101 lb to g (1 kg = 1000 g) %3D g. 65 mph to km/hr (1 km = 0.6214 mi) h. a density of 0.791g/cm3 to Ib/ft3 (1000 g = 2.205 Ib) (1 ft = 30.48 cm) i. a pressure of 14.7 lb/in2 to kg/m² (use a combination of the above)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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