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Diamonds are measured in carats, and 1 carat \(=0.200 \mathrm{g} .\) The density of diamond is 3.51 \(\mathrm{g} / \mathrm{cm}^{3}\) . a. What is the volume of a 5.0 -carat diamond? b. What is the mass in carats of a diamond measuring 2.8 \(\mathrm{mL} ?\)

Short Answer

Expert verified
a. The volume of a 5.0-carat diamond is approximately \(0.285 \ \text{cm}^3\). b. The mass of a diamond with a volume of \(2.8 \ \text{mL}\) is approximately \(49.14 \ \text{carats}\).

Step by step solution

01

Convert the mass of the diamond to grams.

To find the volume, we should first convert the given mass in carats to grams and use the density formula. To do that, multiply the mass in carats by the conversion factor (1 carat = 0.200 g). Mass in grams = (5.0 carats) × (0.200 g / 1 carat) = 1.0 g
02

Calculate the volume of the diamond.

Now that we have the mass of the diamond in grams, it's time to calculate the volume. We can rearrange the density formula to find the volume: V = m / ρ. Volume = Mass / Density V = 1.0 g / (3.51 g/cm³) V ≈ 0.285 cm³ The volume of the 5.0-carat diamond is approximately 0.285 cm³. b. We want to find the carats of a diamond measuring 2.8 mL. Given information: - The diamond's volume is 2.8 mL. - 1 mL = 1 cm³ (since we are working with cm³ in the rest of the problem, we will use this conversion). - The diamond's density is 3.51 g/cm³. - 1 carat = 0.200 g
03

Convert the volume of the diamond to cm³.

We know that the diamond has a volume of 2.8 mL, which is equal to 2.8 cm³ since 1 mL is equal to 1 cm³. So the volume of the diamond is 2.8 cm³.
04

Calculate the mass of the diamond.

Using the density formula, we can find the mass of the diamond. To do so, rearrange the density formula to m = ρ * V. Mass = Density * Volume m = (3.51 g/cm³) * (2.8 cm³) m ≈ 9.828 g The mass of the diamond is approximately 9.828 g.
05

Convert the mass to carats.

Now, we want to find the mass of the diamond in carats. To do that, divide the mass in grams by the conversion factor (0.200 g = 1 carat). Mass in carats = 9.828 g ÷ (0.200 g/carats) ≈ 49.14 carats The mass of the diamond with a volume of 2.8 mL is approximately 49.14 carats.

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Key Concepts

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

Volume
When we talk about volume, we are referring to the amount of space an object occupies. In the exercise, we're asked to find the volume of a diamond. This involves using the density of the diamond and its mass.
To find volume, we use the formula:
  • Volume (\(V\)) = Mass (\(m\)) / Density (\(\rho\))
For our diamond, with a known density, you first convert the carats to grams. Then, you divide by the density to find the volume in cubic centimeters (\(\text{cm}^3\)). This gives us an insight into the diamond's size based on its weight.
Mass conversion
Mass conversion is a crucial step in dealing with diamonds or any scientific calculations involving mass. A diamond's mass is typically given in carats, but for calculations, especially involving density, converting it to grams is necessary.
This conversion involves a simple multiplication:
  • 1 carat = 0.200 grams
In the original exercise, we start with carats and use this factor to obtain the mass in grams. It allows us to plug these values into further formulas, especially when other metrics like volume are involved. Understanding this basic conversion is key to solving related physics and chemistry problems.
Carats
Carats are a unit of weight specifically used for gemstones and pearls. A single carat corresponds to 0.200 grams. This specific measurement is significant in industries like jewelry, where diamond weight can greatly affect value.
To work with diamonds in scientific contexts, carats are often converted to grams using a simple calculation:
  • Carats to grams: Multiply carats by 0.200
  • Grams to carats: Divide grams by 0.200
This kind of conversion ensures that your calculations are accurate, whether you're figuring out volume from weight or the weight based on volume and density.
Understanding carats also helps in comparing gemstones of different sizes and determining the suitability of gemstones for different purposes.
Density formula
The density formula is a fundamental concept in both physics and chemistry. Density (\(\rho\)) is defined as mass (\(m\)) divided by volume (\(V\)):
  • \[ \rho = \frac{m}{V} \]
Density indicates how much mass is contained in a given volume. For example, diamonds have a high density, meaning they're fairly heavy for their size.
This formula proves very useful for calculating either the mass, volume, or density of an object if you know the other two quantities. In the exercise, once we know the diamond's density and either its mass or its volume, we can find the missing value. Using the density formula helps in discerning relationships between these physical properties.

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Most popular questions from this chapter

What data would you need to estimate the money you would spend on gasoline to drive your car from New York to Chicago? Provide estimates of values and a sample calculation.

You have two beakers, one filled to the 100-mL mark with sugar (the sugar has a mass of 180.0 g) and the other filled to the 100-mL mark with water (the water has a mass of 100.0 g). You pour all the sugar and all the water together in a bigger beaker and stir until the sugar is completely dissolved. a. Which of the following is true about the mass of the solution? Explain. i. It is much greater than 280.0 g. ii. It is somewhat greater than 280.0 g. iii. It is exactly 280.0 g. iv. It is somewhat less than 280.0 g. v. It is much less than 280.0 g. b. Which of the following is true about the volume of the solution? Explain. i. It is much greater than 200.0 mL. ii. It is somewhat greater than 200.0 mL. iii. It is exactly 200.0 mL. iv. It is somewhat less than 200.0 mL. v. It is much less than 200.0 mL.

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Explain the fundamental steps of the scientific method.

Perform the following mathematical operations, and express the result to the correct number of significant figures. a. \(6.022 \times 10^{23} \times 1.05 \times 10^{2}\) b. \(\frac{6.6262 \times 10^{-34} \times 2.998 \times 10^{8}}{2.54 \times 10^{-9}}\) c. \(1.285 \times 10^{-2}+1.24 \times 10^{-3}+1.879 \times 10^{-1}\) d. \(\frac{(1.00866-1.00728)}{6.02205 \times 10^{23}}\) e. \(\frac{9.875 \times 10^{2}-9.795 \times 10^{2}}{9.875 \times 10^{2}} \times 100(100 \text { is exact) }\) f. \(\frac{9.42 \times 10^{2}+8.234 \times 10^{2}+1.625 \times 10^{3}}{3}(3 \text { is exact) }\)

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