Solution to Set 4: Difference between revisions
Jump to navigation
Jump to search
| Line 4: | Line 4: | ||
a. In <math>1 m^3</math>, the number of moles is | a. In <math>1 m^3</math>, the number of moles is | ||
<math>{8.885 g\over 1 cm^3} {1 cm^3\over 1.0 × 10^-6 m^3} {1 mol\over 63.55 g} = 1.40 × 10^5 mol\over m^3}</math> | <math>{8.885 g\over 1 cm^3} {1 cm^3\over 1.0 × 10^-6 m^3} {1 mol\over 63.55 g} = 1.40 × 10^5 {mol\over m^3}</math> | ||
Revision as of 17:35, 20 February 2009
Problem 1
Cu, density , atomic mass , fcc structure
a. In , the number of moles is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {8.885 g\over 1 cm^3} {1 cm^3\over 1.0 × 10^-6 m^3} {1 mol\over 63.55 g} = 1.40 × 10^5 {mol\over m^3}}
