105 kg = 4g = ug 0.002ml = .L=. 27 nM = - pM =.. pM 50 pg = 0.25 L = -µl = 20 ml - pl = . pl = ... ..... ..........

Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Mark S. Cracolice, Ed Peters
Chapter3: Measurement And Chemical Calculations
Section: Chapter Questions
Problem 50E: a 0.194 Gg to g, b 5.66 nm to m, c 0.00481 Mm to cm
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105 kg =
ug =
-- ug
0.002ml = .
. L = ...
...
27 nM =
- pM =
*****. ...... pM
***......
****.
50 pg =
g
0.25 L =.
µl = ..
20 ml =.
- pl = ..
pl
..... .
.........
Transcribed Image Text:105 kg = ug = -- ug 0.002ml = . . L = ... ... 27 nM = - pM = *****. ...... pM ***...... ****. 50 pg = g 0.25 L =. µl = .. 20 ml =. - pl = .. pl ..... . .........
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