And thus $rho("density")="mass"/"volume"$. Because the $m^3$ is a very large volume $(1000*L)$, chemists typically use $g=10^-3kg$ to express mass, and $dm^3$ (or $"litres"$) to express volume. $1*dm^-3=(1xx10^-1m)^3=1xx10^-3*m^3=1*L$. Chemists also...
1 Answers 1 viewsLet's break the answer down into three parts: $color (brown) ("First,"$ we're going to convert grams to pounds using this conversion factor: $color(white)(aaaaaaaaaaaaaaaaa$ 1 pound = 453.592 grams $color (purple)...
1 Answers 1 viewsYour goal here will be to use two conversion factors, one to take you from grams to kilograms and one to take you from cubic centimeters to cubic meters....
1 Answers 1 viewsMass is measured in $"kg"$, volume in $"m"^3$ and mass density in $"kg"$ $"m"^-3$. The formula for density is given by the following equation: $rho=m/V$ where $rho$ is the density,...
1 Answers 1 viewsFor a given substance, its tells you the mass occupied by one unit of volume of that substance. In essence, is a measure of how well the molecules of...
1 Answers 1 views$rho,"density"$ $=$ $"Mass"/"Volume"$ Thus, $"volume"$ $=$ $"Mass"/rho$ $=$ $(569*cancelg)/(1.19*cancelg*mL^-1)$ $=$ $??mL$. Note the dimensional consistency. I wanted a volume; and the units of the calculation cancelled to give me an...
1 Answers 1 views$rho="Mass"/"Volume"=(16*g)/(8*cm^-3)$ $=$ $??*g*cm^-3$ And the volume of each piece is $??*cm^3$.
1 Answers 1 views$"Density,"$ $rho$ $=$ $"Mass"/"Volume"$, and we use this quotient. We have $rho$ and $"mass"$. And thus we perform the operation: $"Volume"$ $=$ $"Mass"/rho$ $=$ $(21.0*cancelg)/(19.3*cancelg*mL^-1)$ $=$ $??*mL$. Dimensionally, this is...
1 Answers 1 viewsIf $rho=1.12*g*mL^-1$, then a $1*L$ volume of solution has a mass of: $1*cancelLxx10^3*cancel(mL)*cancel(L^-1)xx1.12*g*cancel(mL^-1)=1.12xx10^3*g$ $-=1.12xxcancel(10^3)*cancelgxxcancel(10^-3)*kg*cancel(g^-1)$ $-=1.12*kg$. Note that this is a measurement SOLELY of $"density"$. This is likely an aqueous solution,...
1 Answers 1 viewsAnd from the initial , we have to first extract the mass of , and then divide this by the combined mass of solutions. And thus we go thru the...
1 Answers 1 views