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A bead weighs 15 g, you place it in a graduated cylinder that has 20mL of water in it. After placing the bead in the cylinder, the water level is now at 30mL. What is its density?
$rho$ 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...
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The water level in a graduated cylinder stands at 20.0 mL before and at 26.2 mL after a 16.74 g metal sample is lowered into the cylinder. What is the density of the sample? What metal is the sample most likely to be?
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.
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An irregularly shaped object that has a mass of 40 g is placed into a graduated cylinder filled with 30 mL of water. The object causes the water to displace to 50 mL. What is the density of the object?
$rho, "density" ="Mass"/"Volume"$. 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.
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A nugget of gold is placed in a graduated cylinder that contains 85 mL of water. The water level rises to 225 mL after the nugget is added to the cylinder. What is the volume of the gold nugget?
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...
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A sample containing 33.42 g of metal pellets is poured into a graduated cylinder initially containing 12.7 mL of water, causing the water level in the cylinder to rise to 21.6 mL. What is the density of the metal?
So we simply take the quotient of mass with respect to the volume of water displaced: $(33.42*g)/((21.6*mL-12.7*mL)$ $=$ $??*g*mL^-1$
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An empty graduated cylinder has a mass of 34.90 g. The cylinder is filled to the 9.40 mL mark with a liquid. The mass the cylinder with liquid is found to be 47.99 g. What is the density of the liquid?
$"Density"$ $=$ $"Mass"/"Volume"$ $=$ $(47.99*g-34.90*g)/(9.40*mL)$ $=$ $??g*mL^-1$.
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A 147-g piece of metal has density of 7.00 g/mL. A 50-mL graduated cylinder contains 20.0 mL of water. What is the final volume after the metal is added to the graduated cylinder?
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...
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A graduated cylinder contains 225 mL of water. What is the new water level affer 30.2 g of silver metal with a density of 10.5 g/mL is submerged in the water?
Well, we work out the water displaced by the silver........by calculating the volume of the silver nugget using the . $"Density"(rho)$ $=$ $"Mass"/"Volume"$ Thus $"Volume"="Mass"/rho$ $=$ $(30.2*g)/(10.5*g*mL^-1)$ $~=$ $3*mL$. This...
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A graduated cylinder has a mass of 50 g when empty. When 30 mL of water is added, it has a mass of 120 g. If a rock is added to the graduated cylinder, the level rises to 75 ml and the total mass is now 250 g. What is the density of the rock?
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...
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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?
Do 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?
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