The density of pure silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). If \(5.25 \mathrm{~g}\) of pure silver pellets is added to a graduated cylinder containing \(11.2 \mathrm{~mL}\) of water, to what volume level will the water in the cylinder rise?

Short Answer

Expert verified
The new water level in the graduated cylinder after adding the silver pellets will be 11.7 mL.

Step by step solution

01

Understand the relation between mass, volume, and density

Density is given by the equation: \[Density = \frac{Mass}{Volume}\] We can rearrange this formula to find the volume: \[Volume = \frac{Mass}{Density}\]
02

Find the volume of the silver pellets

Now we will apply the formula to find the volume of the silver pellets. The mass is given as 5.25g, and the density is given as 10.5 g/cm³. Plugging these values into the formula, we get: \[Volume_{silver} = \frac{Mass_{silver}}{Density_{silver}} = \frac{5.25 g}{10.5 \frac{g}{cm^{3}}}\] Calculating this value, we get: \[Volume_{silver} = 0.5 cm^{3}\] So, the volume of the silver pellets is 0.5 cm³.
03

Determine the volume displacement

When we add the silver pellets into the graduated cylinder filled with water, the silver pellets will displace a volume of water equal to their own volume. In other words, the water level will rise by the same volume of silver that we introduced.
04

Calculate the new water level

Initially, we have 11.2 mL of water in the graduated cylinder. Since 1 mL is equal to 1 cm³, we have 11.2 cm³ of water initially. After adding the silver pellets, the water level will rise by 0.5 cm³. Therefore, the new volume level can be calculated as: \[Volume_{new} = Volume_{water} + Volume_{silver} = 11.2 cm^{3} + 0.5 cm^{3}\] Finally, we find: \[Volume_{new} = 11.7 cm^{3}\] Thus, the new water level in the graduated cylinder will be 11.7 mL (since 1 cm³ is equal to 1 mL).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass to Volume Ratio
Understanding the mass to volume ratio is an essential step in many scientific calculations, particularly when working with the concept of density. Density is a measure of how much mass is contained in a given volume, and the mass to volume ratio is essentially another way to express this relationship.

Let's break this down: if you have a certain amount of substance, say silver, knowing its mass to volume ratio lets you determine how much space it will take up. For instance, pure silver has a known density, which tells us its mass to volume ratio. When you have 5.25 g of pure silver, you can use its density to calculate the volume this mass of silver would occupy.

In practical situations, the mass to volume ratio lets you predict the outcome when adding a solid to a liquid, such as putting pellets into water. It answers the question, 'How much will the level of the liquid rise?' By applying the formula discussed in the exercise, you can predict this change without needing to physically conduct the experiment, which can save time and resources.
Volume Displacement Method
The volume displacement method is a practical approach to determine the volume of an irregularly shaped object, like silver pellets. This method is based on the principle that when an object is submerged in water, it displaces an amount of water equal to its volume.

The classic school experiment demonstrates this beautifully: a graduated cylinder is partially filled with water, and the initial volume is noted. When an object is added, the water level rises — this increase is the volume of the object. In the given example, by dropping 5.25 g of silver into the water, we cause the liquid to display this displaced volume, allowing us to calculate the change in water level accurately.

Applying the Volume Displacement Method

By strictly adhering to the proper technique, you can ensure accurate results when using the volume displacement method. Place the object gently into the liquid to avoid splashing or additional displacement, and ensure the object is fully submerged for an accurate measurement. The simplicity and effectiveness of this method make it a staple in scientific studies and educational demonstrations.
Density Formula Application
When it comes to determining the density of an object, or its mass to volume ratio, we have a straightforward formula at our disposal: \[Density = \frac{Mass}{Volume}\].This fundamental formula allows us to calculate one variable, as long as the other two are known. If you want to find the volume, you can reorganize the density formula as done in the solution: \[Volume = \frac{Mass}{Density}\].Applying the density formula is crucial in many scientific and real-world applications. In the example of silver pellets, knowing the density (mass to volume ratio) of silver and the mass of the pellets, you can calculate the volume they'll occupy. This calculation tells us the exact increase in water level when the pellets are submerged in the graduated cylinder.

Understanding how to manipulate and apply the density formula is beneficial, not only in solving textbook problems but also in real-life situations such as crafting jewelry or designing objects that must have specific buoyancy characteristics. This exercise demonstrates the practical application of the formula, providing an invaluable tool for predicting the outcomes in various scenarios involving mass, volume, and density.

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Most popular questions from this chapter

Perform the following mathematical operations, and express the result to the correct number of significant figures. a. \(\frac{2.526}{3.1}+\frac{0.470}{0.623}+\frac{80.705}{0.4326}\) b. \((6.404 \times 2.91) /(18.7-17.1)\) c. \(6.071 \times 10^{-5}-8.2 \times 10^{-6}-0.521 \times 10^{-4}\) d. \(\left(3.8 \times 10^{-12}+4.0 \times 10^{-13}\right) /\left(4 \times 10^{12}+6.3 \times 10^{13}\right)\) e. \(\frac{9.5+4.1+2.8+3.175}{4}\) (Assume that this operation is taking the average of four numbers. Thus 4 in the denominator is exact.) f. \(\frac{8.925-8.905}{8.925} \times 100\) (This type of calculation is done many times in calculating a percentage error. Assume that this example is such a calculation; thus 100 can be considered to be an exact number.)

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