A foam material has a density of $175$ $g$$/L$. What is its density in units of $lb$$/ft^3$?

Suppose if $v$ is the relative velocity between two frames $X$ (earth) and $X'$ (quasar). The relative velocity of an object in the two frames are related to each other...

Your 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....

The radius= 1.08cm = 0.43in $density=(mass)/(volume)$ $Volume = (mass)/(density)$ $Volume=(82)/(2.7)=30.37cm^3$ The volume of a sphere is given by: $V =4/3pir^3$ So $ r^3=(3V)/(4pi) =(3xx30.37)/(4xx3.142)=1.274$ $r = 1.084cm$

$"Molar mass (Cu)"$ $=$ $63.55$ $g*mol^-1$. $"Molar mass (Pt)"$ $=$ $195.08$ $g*mol^-1$. This threefold difference in molar mass, of course translates to . Given that the difference in densities is...

By definition, $rho,"density"$ $=$ $"Mass of stuff"/"Volume of stuff"$. We want to find a volume, so: $"Volume"$ $=$ $"Mass"/rho$ $=$ $(175*cancelg)/(13.6*cancelg*mL^-1)$. Note that this calculation is dimensionally consistent. We want...

Let the volume of the body be $vcm^3$ the of the body be $dgcm^-3$ So weight of the body in air$W_"air"=vdg$ dyne, $"where "g->"acceleration due to gravity"$ Now the bouyant...

In your example, $rho=("Mass")/("Volume")=(598*g)/(1.2*cancelLxx10^(3)*mL*cancel(L^-1))$ $=$ $0.498*g*mL^-1$. Is the solution likely to be aqueous? Why or why not?

Well we can work out the density..... $rho="Mass"/"Volume"=(8.184*g)/(10.5*mL-1.2*mL)=0.88*g*mL^-1$. This could literally be anything.........

Soaps form foam very easily with soft water. But they first form lather in hard water and then form foam which takes time. Soaps take time to neutralise the calcium...

To find the number of formula units in 0.175 mol of ammonium sulfate solid, you can simply multiply the number of mole by Avogadro's number : $?(NH_4)_2SO_4=0.175cancel(mol)xx(6.022xx10^23" formula units")/(1cancel(mol))=1.05xx10^23" formula...