A graduated cylinder contains 155 mL of water. A 15.0-g piece of ...

Question

A graduated cylinder contains 155 mL of water. A 15.0-g piece of iron and a 20.0-g piece of lead are added. What is the new water level, in milliliters, in the cylinder(see TABLE2.8)? (2.7)

Accepted Answer

Welcome back everyone. The initial water level in a graduated cylinder is 75. mL. Two metals are added to the graduated cylinder. Copper and antimony weighing 18.9 g and 25.3 g respectively, determine the new water level in milliliters in the graduated cylinder. The densities of copper and antimony are 8.96 g per milliliter and 6.68 g per milliliter respectively. So we have our graduated cylinder and let's just draw one out really quick. And we're told that our initial water level Is around 75.2 ml. So this is, we'll say VI. Now we have a piece of copper and antimony which I'll express by these two circles here. And these two metals are both added into our graduated cylinder which raises our water level to a new volume, which we'll call V F. And we need to figure out what this new volume would be. Recall that copper has a chemical symbol of C U and antimony has a chemical symbol of S B. Now, in order to determine our final volume, we should write out an expression where we can determine that V F our final water level is going to be equal to the initial volume level of our water in the cylinder plus our volume of copper plus our volume of antimony. And because we're given the densities and masses of our metals, copper and antimony, we're going to want to recall our formula to calculate density represented by row where density is equal to mass divided by volume. And in order to isolate for volume, we're going to multiply both sides by volume to cancel volume out on the right hand side, which would give us density times volume equal to mass. And then isolating for volume, we're going to divide both sides by row so that we have our volume equal to mass divided by row. And so this is how we will find the individual volumes of our metals. So to begin with finding the volume of copper, we're going to use its given mass from the prompt as 18. g and divide by its given density from the prompt as 8.96 g per milliliter. And so canceling out grams were left with milliliters as our final unit. And in our calculators, we'll come up with a volume for copper as 2. mL. And actually, we can round this to 2.1094 ml to make things simpler. Now, let's find our volume of antimony Recall that this is set equal to its mass given in the prompt as 25.3 g in the numerator and divided by our given density of antimony from the prompt as 6.5, 68 g per milliliter in our denominator. So again, canceling out grams, we're left with milliliters as our unit of volume for antimony. And we'll come up with a volume as 3.7874 mL for antimony's volume. Now, with our volumes of our metals calculated, we can finally get to that term V F to solve To get our final answer. So this is set equal to our given initial volume for the water in the cylinder as 75.2 ml. This is added to our volume of copper which we found above as 2.1094 milliliters and added to our volume of antimony, which above we found to be 3. mL. And so taking the sum between all of our terms here, we're coming up with our final volume in the cylinder as 81.1 mL. And note that I used three sigfig for our final answer because based on all of our given values for our volume and masses as well as even our densities, all of our terms have three significant figures. And so our final answer should be with three significant figures. And so this would actually complete this example as the final Water level in our graduated cylinder after copper and antimony are added. And so this will correspond to choice D which states that we have a volume of 81.1 ml. I hope that this made sense and let us know if you have any questions.

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A graduated cylinder contains 155 mL of water. A 15.0-g piece of iron and a 20.0-g piece of lead are added. What is the new water level, in milliliters, in the cylinder(see TABLE2.8)? (2.7)

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