## Chemistry 9th Edition

To find the mass of CO in the room that would register 100 ppm on a carbon monoxide detector, use dimensional analysis as follows: $\frac{(18\times12\times8)ft^{3}}{1}\times(\frac{(12)in}{(1)ft})^{3}\times(\frac{(2.54)cm}{(1)in})^{3}\times(\frac{(1)m}{(100)cm})^{3}\times\frac{(400000)ug}{(1)m^{3}}\times\frac{(1)g}{(1000000)ug} \approx 19.573g$ This is $20g$ to the correct number of significant figures (1).