A graduated cylinder is filled with water to a level of 40.0 ml. When a piece of copper is lowered into the cylinder, the water level rises to 63.4 mL. What is the volume of the copper sample? If the density of the copper is 8.9 g/cm^, what is its mass?

$rho$ $=$ $(28*g)/(5*mL)$ $=??g*mL^(-1)$

That metal occupies a volume of 6.2 mL. Its is $16.74/6.2=2.7 g cm^-3$. Consulting I could say that the metal is aluminium.

$rho$ $=$ $"Mass"/"Volume displaced"$ $=$ $(8.25*g)/((26.47-21.25)*mL)$ $=$ $??g*mL^-1$

The water level rose from $"45 ml"$ to $"55.5 ml"$, so that is a volume increase of $"10.5 ml"$. Assuming that the density of the water is $"1.00 g/ml"$, then...

A more interesting question is the value of the nugget (more interesting to me anyway). As a heavy metal, gold is incredibly dense, and has $rho=19.3*g*mL^-1$; this is almost twice...

So we simply take the quotient of mass with respect to the volume of water displaced: $(33.42*g)/((21.6*mL-12.7*mL)$ $=$ $??*g*mL^-1$

$"Density"$ $=$ $"Mass"/"Volume"$ $=$ $(47.99*g-34.90*g)/(9.40*mL)$ $=$ $??g*mL^-1$.

Note: $1 " ml" = 1 " cc"$ If the water rose $7" ml" - 2" ml"=5" ml"$ then the stone had a volume of $5" ml" or 5 "...

Use the formula to determine the volume of the piece of metal. $"density"="mass"/"volume"$ Rearrange the equation to isolate volume. $"volume" ="mass"/"density"$ $"volume"=(147 cancel"g")/(7.00cancel"g"/"mL")="21.0 mL"$ The final volume in the cylinder...

We're asked to find the of the rock, given some mass and volume information. We're given that the mass of the system before the rock was added was...