NAME CHEM 103 TEST IA ECC PROFONWUNAKA I. Match each value in the left-hand colu
ID: 546923 • Letter: N
Question
NAME CHEM 103 TEST IA ECC PROFONWUNAKA I. Match each value in the left-hand column with its expression in scientific notation in the right-hand column. 1) 0.0201 _.. 2) 14,916 14916 x 10 b) 2.01 x 10 )7.50 x 10 d) 2.01x 10 e) 1.4916 x 10 D 7.50x 10 3) 0.750 x 10 4 0.0000750 5) 14.916 x 102 _ 6) 201 2. An irregularly shaped object has a mass of 2.3 x 10*g and a density of 3.4 x 10 kg/m The volume of this object would be calculated to be xleg x 2.3 x10-2 g 2.3 x 10g x 2.3 x 10g 3.4 x 10 kg 10 g x 2.3 x 102g 3.4 x 10'kg 10'g e) none of the above 3. Make the following conversions. Observe significant figures. d) 91.2°F to °C a) 6.45 x 10 g to pounds b) 3.2 x 10 m to inches e) 14.7 ibi to gr c) 4.5 gal to LExplanation / Answer
1. 1) 0.0201 = b) 2.01 x 10-2
2) 14.916 = a) 1.4916 x 101
3) 0.750 x 104 = f) 7.50 x 103
4) 0.0000750 = c) 7.50 x 10-5
5) 14.916 x 10-5 = e) 1.4196 x 10-4
6) 201 = d) 2.01 x 102
2. Given , mass = 2.3 x 10-2 g = 2.3 x 10-2 x 1 /1000 kg = 2.3 x 10-5 kg and density = 3.4 x 103 kg/m3
We have the relation as, density = mass / volume
=> volume = mass / density
=> volume = 2.3 x 10-5 kg / 3.4 x 103 kg /m3
Therefore the volume of the object can be calculated as---
d) (m3 / 3.4 x 103 kg) * ( 1 kg / 103 g) * (2.3 x 10-2 g)
3.
a) 6.45 x 104 g to pounds
(6.54 x 104 g) * (1 pound / 453.592 g) = 65400 / 453.592 pounds = 144.18 pounds
b) 3.2 x 102 m to inches
(3.2 x 102 m) * (1 inch / 0.0254 m) = 320 / 0.0254 inch = 12598.43 inches
c) 4.5 gal to L
(4.5 gal ) * ( 1 L / 0.264172 gal) = 4.5 /0.264172 L = 17.03 L = 17 L
d) 91.20F to 0C
We have the formula as-
Temperature on Fahrenheit scale =( 9/5 x C ) + 32
=> 91.2 = (9/5 x C ) + 32
=> 91.2 -32 = 9/5 X C
=> 59.2 = 9/5 x C
=> 296 /9 = C
=> C = 32.89
Therefore, the required tmperature in 0C = 32.89
e) 14.7 lb/in2 to g/m2
14.7 lb / in2 x (1000 g / 2.205 lb) x (1 in2 / (2.54cm)2) x ((100cm)2 / 1m2) = 1.03 x107 g/m2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.