A 50.0-g silver object and a 50.0-g gold object are both added to...

Question

A 50.0-g silver object and a 50.0-g gold object are both added to 75.5 mL of water contained in a graduated cylinder. What is the new water level, in milliliters, in the cylinder(see TABLE2.8)? (2.7)

Accepted Answer

Welcome back everyone. A 25 g piece of silver and a 25 g piece of tungsten are submerged in a graduated cylinder containing water. The initial water level is 59.7 millilitres, determine the new water level in milliliters in the graduated cylinder. After the two metals are added, the densities of silver and tungsten are 10.5 g per milliliter and 19.3 g per milliliter respectively. So we'll draw out a sketch of a graduated cylinder as follows. And we are told that initially the water level is equal to .7 ml. So this is p initial as our initial volume. And this line that we've drawn in for this volume is our meniscus of that water level. Now, then we're told we have a piece of tungsten and recall that the chemical symbol for tungsten is a capital W and we have also a piece of silver, which it is represented by the chemical symbol A G for silver that are added into our graduated cylinder, which will then elevate our volume of our water inside of the graduated cylinder to a higher level which we call V F and this is what we need to figure out our final volume of our water. And so we're told to find this in milliliters now because we are given the densities of our metals, tungsten and silver. We're going to recall our formula which relates row for density equal to mass divided by volume. And so we want to isolate to solve for our volume, meaning we're gonna multiply both sides by our denominator V so that it cancels out on the right, we'll have row times volume equal to mass and then to isolate volume. We're going to divide both sides by row. That's going to show us that we have volume equal to mass divided by row. Now, we can utilize this formula to find our volumes of our two metals, which would be important because we should recognize that in order to get the final volume of our water, we're going to want to take the initial volume level of the water, add it to our volume of our metal silver, add it to our volume of tungsten. And so next, what we'll have is for our volume of silver, we would have its mass given in the prompt as 25.0 g in the numerator divided by its density given as 10.5 g per milliliter. Our units of grams will cancel leaving us with milliliters. And this will give us a volume for silver equal to 2.3810 mL. Next, we solve for our volume of tungsten equal to the given molecular or rather the given mass of tungsten being 25.0 g as well divided by its given density of 19.3 g per milliliter. And just as before our units of grams cancel were left with milliliters. And we get a volume equal to 1.2953 mL. Now with our two volumes of our metals, we're just going to plug in to find the final volume of our water when the metals are added to the cylinder. So we have the initial volume given as 59.7 millilitres plus our volume with silver of 2.3810 mL plus our volume with tungsten being 1. mL. Now taking the sum of each of these volumes, we'll find a final volume for the water in the cylinder equal to 63 .3763 ml in which we want to round based on our given values. And the prompt to the least precise measurement as three significant figures which will give us our final answer. As 63.4 ml being the final volume of our water within the graduated cylinder. Once our metals tungsten and silver are added inside. So 63.4 ml is our final answer. I hope this made sense and let us know if you have any questions

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A 50.0-g silver object and a 50.0-g gold object are both added to 75.5 mL of water contained in a graduated cylinder. What is the new water level, in milliliters, in the cylinder(see TABLE2.8)? (2.7)

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