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An irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 2 mL. The height of the water rose to 7 mL. If the mass of the stone was 28 g, what was its density?
$rho$ $=$ $(28*g)/(5*mL)$ $=??g*mL^(-1)$
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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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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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An irregularly shaped stone was lowered into a graduated cylinder holding a of volume of water equal to 2 mL. The height of the water rose to 7 mL. If the mass of the stone was 25 g, what was its density?
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 "...
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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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An empty graduated cylinder weighs 45.8772 g. After 20.0 mL of a liquid is added, the cylinder and its contents weigh 77.7572 g. What is the density of the liquid in g/mL?
Given, $V = 20.0"mL"$ $m = 77.7572"g" - 45.8772"g" approx 31.880"g"$ Recall, $rho = m/V$ Hence, $rho = m/V approx (1.59"g")/"mL"$
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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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