An irregular object with a mass of 12 kg displaces 1.75 water when placed in an overflow container. How do you calculate the density of the object?

Fill a measuring cylinder to a known volume of water, say $V_0$. Place the solid object of unknown volume $V_s$ in the water and ensure it is totally submerged. Read...

$rho, "density" ="Mass"/"Volume"$. The object displaces a volume of $(50-30)*mL =20*mL$. Given that we know the mass the calculation of the density is now easy.

Clearly, the object has displaced $15*mL$ of water, and thus has a volume of $15*mL$. $rho", density"$ $=$ $(45*g)/(15*mL)$ $=??*g*mL^-1$.

$rho=m/V=(45g)/(15cm^3)=3g//cm^3$

$"density"=("mass")/("volume")$ The $"60 mL"$ of displaced water is the volume of the PVC pipe. Since the PVC pipe is a solid, we can use the equality $"1 mL"$$=$$"1 cm"^3"$ to...

$rho, "density"="Mass"/"Volume"$, by definition. And thus we simply take the quotient: $(1450*g)/(542*mL)=??*g*mL^-1$

$rho,"density"="mass"/"volume"$ And thus $19.3*g*cancel(cm^-3)xx0.950*cancel(cm^3)="mass of the nugget"$ $=$ $18.3*g$. The jeweller would offer $18.3*gxx$8=00*g^-1~=$150-00$. The current spot price for gold is $"39-00 USD"*g^-1$. I would tell that jeweller that he...

$d=$? $m=262.2" g"$ The volume of the mineral is equal to the volume of water it displaced, because the of water is $1" g"//"cm"^3$ $V=46 " cm"^3$ Formula for density...

And thus we work out the quotient with the given data; note that we have to subtract the tare weight, i.e. the mass of the container: $"Mass"/"Volume"=(1300*g-250*g)/(1000*mL)=(1050*g)/(1000*mL)$ $=1.05*g*mL^-1...................$

The equivalent mass of a metal $"M"$ is the mass that will displace 1.008 g of $"H"$ (one-half mole of $"H"_2$). At STP, (0 °C and 1 bar) the is...