$rho$ $=$ $(28*g)/(5*mL)$ $=??g*mL^(-1)$
1 Answers 1 views$rho$ is equal to $""mass"/"volume"$ so in our case the mass is $15g$ while the volume should be: $30-20=10ml=10cm^3$ The density of the material of the bead should then...
1 Answers 1 viewsA 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...
1 Answers 1 viewsSo we simply take the quotient of mass with respect to the volume of water displaced: $(33.42*g)/((21.6*mL-12.7*mL)$ $=$ $??*g*mL^-1$
1 Answers 1 viewsFor your problem, $rho=((50.92-25.23)*g)/(25.0*mL)$ $=$ $1.03*g*mL^-1$ Alternatively $rho=((50.92-25.23)*g)/(25.0xx10^-3L)$ $=$ $??g*L^-1$
1 Answers 1 views$"Density"$ $=$ $"Mass"/"Volume"$ $=$ $(47.99*g-34.90*g)/(9.40*mL)$ $=$ $??g*mL^-1$.
1 Answers 1 viewsNote: $1 " ml" = 1 " cc"$ If the water rose $7" ml" - 2" ml"=5" ml"$ then the stone had a volume of $5" ml" or 5 "...
1 Answers 1 viewsUse 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...
1 Answers 1 viewsWe'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...
1 Answers 1 viewsDo you agree? The copper displaces the given volume of water. Now $rho_"Cu"=8.90*g*cm^3$ OR $rho_"Cu"=8.90*g*mL^-1$, i.e. $1*mL-=1*cm^3$ But by definition, $rho_"density"="mass"/"volume"$ And thus $"mass"=rhoxx"volume"=8.90*g*cancel(mL^-1)xx23.4*cancel(mL)$ $=208.3*g$. Do you follow?
1 Answers 1 views