Chapter 5 Review HW = Study for tomorrow’s exam (bring your calculator Converting between two sets of units never changes the number of significant figures in a measurement. Remember, data are only as good as the original measurement, and no later manipulations can clean them up. “No human endeavor can be called science if it can not be demonstrated mathematically.” Leonardo da Vinci (1452-1519)

A physical quantity must include: A number and a ???

A physical quantity must include: Number + Unit

Read the thermometer correctly… (Celcius) 100 50

Read the thermometer correctly… (Celcius) 100 50 40. oC

How many kilometers is 15,000 decimeters?

How many kilometers is 15,000 decimeters? ( ) ____ ( ) ______ 1 m 10 dm 1,000 m 1 km 15,000 dm 1.5 km =

How many seconds is 4.38 days?

How many seconds is 4.38 days? ____ ( ) ( ) _____ ( ) ____ 24 h 1 d 1 h 60 min 1 min 60 s 4.38 d = 3.78 x 105 s

Measured dimensions of a rectangular solid: Length = 15.2 cm Width = 3.7 cm Height = 8.6 cm H W Find volume of solid in m3 L V = L . W . H = (15.2 cm)(3.7 cm)(8.6 cm) = 480 cm 3

( ) ( ) ( ) 0.000480 m3 4.80 x 10-4 m3 Convert 480. cm3 to m3 _____ cm.cm.cm 0.000480 m3 = 4.80 x 10-4 m3 =

If light travels at the speed of 2.998 x108 meters per second, at how many kilometers per hour does it travel ?

( ) ( ) ( ) How many km/hr is 2.998 x 108 m/s? 1.079 x 109 km/hr 60 s ____ ( ) 60 min 1 min 60 s m s 2.998 x 108 1 hr 1000 m = 1.079 x 109 km/hr

An irregularly-shaped sample of aluminum (Al) is put on a balance and found to have a mass of 43.6 g. The student decides to use the water-displacement method to find the volume. The initial volume reading is 25.5 mL and, after the Al sample is added, the water level has risen to 41.7 mL. Find the density of the Al sample in g/cm3. (Remember: 1 mL = 1 cm3.)

Volume of Al = Final Volume – Initial Volume mass = 43.6 g final volume = 41.7 mL initial volume = 25.5 mL density = ? g/cm3 Volume of Al = Final Volume – Initial Volume = 41.7 mL – 25.5 mL = 16.2 mL D = M = 43.6 g = 2.69 g/mL = 2.69 g/cm3 V 16.2 mL

Find the volume that 35.2 g of carbon tetrachloride (CCl4) will occupy if it has a density of 1.6 g/mL.

mass = 35.2 g density = 1.6 g/cm3 volume = ? mL V = M = 35.2 g = 22 cm3 = 22 mL D 1.6 g . cm3