If the density of a material is $0.88*g*mL^-1$, what is the material?

Let'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)...

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$ $r = 1.084cm$

Density, $rho$, $=$ $(37.72*g)/(6.80*cm^3)$ $=$ $??$ $ g*cm^(-3)$

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

The equation for density is mass divided by volume i.e. $D = m/V$. The mass of the material is $5.6$ $g$. The volume is $2.2$ $mL$. If you're wondering why,...

$"Density"$ $=$ $"mass"/"volume"$ $=$ $(8*g)/(2*cm^3)$ $=$ $4*g*cm^-3$ When we put the units into these sorts of calculations, it is an extra check on our accuracy. We try to calculate a...

And thus $"Volume"="Mass"/rho$ $=$ $(3.50*cancelg)/(2.98*cancelg*mL^-1)$ And thus $"volume"$ $>$ $1.0*mL$, because $1/(mL^-1)$ $-=$ $"mL"$. So we know we;ve got the right units for the calculation.

And so $"mass"=rhoxx"volume"=6.72*g*cancel(cm^-3)xx55.1*cancel(cm^3)$ $-=370.3*g$

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...

$"Density"$ $=$ $"Mass"/"Volume"$ I have an idea that the of silver metal is greater than $10.0*g*mL^-1$ (this is typical of the heavier metals!). You will have to look this up...