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Gold can be hammered into extremely thin sheets called gold leaf. If a \(200-\mathrm{mg}\) piece of gold (density \(=19.32 \mathrm{~g} / \mathrm{cm}^{3}\) ) is hammered into a sheet measuring \(2.4 \times 1.0 \mathrm{ft}\), what is the average thickness of the sheet in meters? How might the thickness be expressed without exponential notation, using an appropriate metric prefix?

Short Answer

Expert verified
The average thickness of the gold leaf is approximately 4.651 x 10鈦宦孤 m or 0.4651 nm (nanometers).

Step by step solution

01

Convert mass of gold to grams

The initial mass of gold is given in milligrams (mg). First, let's convert it to grams (g). Given mass = 200 mg 1 g = 1000 mg Therefore, mass of gold in grams = 200 mg / 1000 = 0.2 g
02

Calculate volume of the gold leaf

Now we need to find the volume of the gold leaf using density and mass. Density is mass per unit volume, so we use the formula: Volume = mass/density The mass of gold is 0.2 g, and the density is 19.32 g/cm鲁. Volume = 0.2 g / 19.32 g/cm鲁 鈮 0.01035 cm鲁
03

Convert the area of the gold leaf to square meters

Next, we'll convert the dimensions of the gold leaf into meters and calculate the area. The dimensions are given in feet: 2.4 ft x 1.0 ft. 1 ft = 0.3048 m So, dimensions in meters = 2.4 ft x 0.3048 m/ft x 1.0 ft x 0.3048 m/ft 鈮 0.73152 m x 0.3048 m Area = 0.73152 m x 0.3048 m 鈮 0.2229 m虏
04

Calculate the thickness of the gold leaf

Now that we have the volume and area, we can calculate the thickness using the formula: Thickness = Volume/Area First, convert the volume to cubic meters: Volume = 0.01035 cm鲁 x (0.01 m/cm)鲁 鈮 1.035 x 10鈦宦孤 m鲁 Now, calculate the thickness: Thickness = (1.035 x 10鈦宦孤 m鲁) / (0.2229 m虏) 鈮 4.651 x 10鈦宦孤 m
05

Express the thickness without exponential notation using an appropriate metric prefix

The thickness of the gold leaf is 4.651 x 10鈦宦孤 m. We can express the thickness without exponential notation using nanometers (nm): 1 m = 10鈦 nm Therefore, the thickness in nanometers = 4.651 x 10鈦宦孤 m x 10鈦 nm/m 鈮 0.4651 nm

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Key Concepts

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

Density and Volume Calculations
Understanding density and volume calculations is essential for tasks like measuring the size of a gold leaf, or any other object where you have to determine the amount of space it occupies. When you're dealing with materials like gold, which can be shaped into very thin sheets, the concept becomes even more crucial.

Let's start with density, which is defined as the mass of a material per unit volume. It's typically expressed using the formula: \[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]. In our exercise, gold's density is stated as 19.32 grams per cubic centimeter (g/cm鲁), and we have a mass of 0.2 grams. To find the volume that this mass of gold would take up, we rearrange the formula to solve for volume, leading to \[ \text{Volume} = \frac{\text{mass}}{\text{density}} \].

In practice, when you're working these kinds of problems, always ensure the units for mass and volume are compatible with the density units provided. For gold with a given density, calculating the volume tells you how much space the gold takes up, which is a small but crucial step towards finding out how thin the gold leaf can be.
Unit Conversion
Unit conversion is a pivotal skill in physics and chemistry problems, as it allows you to work seamlessly with measurements in different units. Since the metric system and other systems, like the imperial system, use different base units, being able to convert between these units is essential.

In the exercise on gold leaf thickness, we encounter two instances where unit conversion is required. First, when converting the mass of gold from milligrams to grams. The basic conversion here is \(1 \text{g} = 1000 \text{mg}\), which allows us to convert 200 mg to 0.2 g. The second instance involves converting the measurements of the gold leaf from feet to meters, knowing that \(1 \text{ft} = 0.3048 \text{m}\), a critical step to determine the gold leaf's area in square meters.

Unit conversion typically involves multiplication or division by a conversion factor, which is a ratio that represents how a measurement in one unit relates to the measurement in another unit. By doing so, we preserve the equality (since the conversion factor is essentially '1') and correctly transform our quantity into the desired units.
Metric Prefixes
Metric prefixes are incredibly helpful in science, as they yield a way to express very large or very small numbers in a more digestible format, using the powers of ten. When dealing with dimensions like the thickness of gold leaf which can be tremendously small, using metric prefixes like 'nano' for nanometers, makes the numbers easier to read and understand.

The metric system has a set of prefixes that stand for certain powers of ten. For example, 'kilo-' means a thousand or \(10^3\), 'milli-' means a thousandth or \(10^{-3}\), and 'nano-' signifies a billionth or \(10^{-9}\). In our gold leaf exercise, converting the thickness from meters to nanometers鈥擻(1 \text{m} = 10^{9} \text{nm}\)鈥攎akes the final result more tangible: from about \(4.651 \times 10^{-11} \text{m}\) to approximately 0.4651 nanometers. Understanding and using metric prefixes allows us to communicate measurements clearly without resorting to complex exponential notation, which can be particularly daunting when the numbers are very small or very large.

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Most popular questions from this chapter

The Morgan silver dollar has a mass of \(26.73 \mathrm{~g}\). By law, it was required to contain \(90 \%\) silver, with the remainder being copper. (a) When the coin was minted in the late 1800s, silver was worth \(\$ 1.18\) per troy ounce (31.1 g). At this price, what is the value of the silver in the silver dollar? (b) Today, silver sells for about \(\$ 13.25\) per troy ounce. How many Morgan silver dollars are required to obtain \(\$ 25.00\) worth of pure silver?

In 2006, Professor Galen Suppes, from the University of Missouri-Columbia, was awarded a Presidential Green Challenge Award for his system of converting glycerin, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\), a by-product of biodiesel production, to propylene glycol, \(\mathrm{C}_{3} \mathrm{H}_{6}(\mathrm{OH})_{2}\). Propylene glycol produced in this way will be cheap enough to replace the more toxic ethylene glycol that is the primary ingredient in automobile antifreeze. (a) If \(50.0\) mL of propylene glycol has a mass of \(51.80 \mathrm{~g}\), what is its density? (b) To obtain the same antifreeze protection requires \(76 \mathrm{~g}\) of propylene glycol to replace each \(62 \mathrm{~g}\) of ethylene glycol. Calculate the mass of propylene glycol required to replace \(1.00\) gal of ethylene glycol. The density of ethylene glycol is \(1.12 \mathrm{~g} / \mathrm{mL}\). (c) Calculate the volume of propylene glycol, in gallons, needed to produce the same antifreeze protection as \(1.00\) gallon of ethylene glycol.

The distance from Earth to the Moon is approximately 240,000 mi. (a) What is this distance in meters? (b) The peregrine falcon has been measured as traveling up to \(350 \mathrm{~km} / \mathrm{hr}\) in a dive. If this falcon could fly to the Moon at this speed, how many seconds would it take?

Suppose you are given a sample of a homogeneous liquid. What would you do to determine whether it is a solution or a pure substance?

Two students determine the percentage of lead in a sample as a laboratory exercise. The true percentage is \(22.52 \%\). The students' results for three determinations are as follows: 1\. \(22.52,22.48,22.54\) 2\. \(22.64,22.58,22.62\) (a) Calculate the average percentage for each set of data, and tell which set is the more accurate based on the average. (b) Precision can be judged by examining the average of the deviations from the average value for that data set. (Calculate the average value for each data set, then calculate the average value of the absolute deviations of each measurement from the average.) Which set is more precise?

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