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At room temperature the element bromine, \(\mathrm{Br}_{2},\) is a liquid with a density of 3.12 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate the mass of 125 \(\mathrm{mL}\) of bromine. What volume does 85.0 \(\mathrm{g}\) of bromine occupy?

Short Answer

Expert verified
The mass of 125 mL of bromine is 390 g, and the volume occupied by 85.0 g of bromine is approximately 27.2 cm³.

Step by step solution

01

Convert volume to an appropriate unit.

The given volume is in mL, but the units for density are g/cm³. So, the volume should also be in cm³. We can convert the volume to cm³ using the conversion factor 1 cm³ = 1 mL: \(V = 125\,\text{mL} \times \frac{1\,\text{cm}^{3}}{1\,\text{mL}} = 125\,\text{cm}^3\)
02

Calculate the mass of bromine.

Using the given density and volume, we can calculate the mass using the formula: Mass = Density × Volume \(m = 3.12\,\frac{\text{g}}{\text{cm}^3} \times 125\,\text{cm}^3\) \(m = 390\,\text{g}\) So, the mass of 125 mL of bromine is 390 g.
03

Calculate the volume of 85.0 g of bromine.

Using the given density and mass, we can calculate the volume using the formula: Volume = Mass / Density \(V = \frac{85.0\,\text{g}}{3.12\,\frac{\text{g}}{\text{cm}^{3}}}\) \(V \approx 27.2\,\text{cm}^{3}\) So, the volume occupied by 85.0 g of bromine is approximately 27.2 cm³.

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Key Concepts

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

Volume Conversions
When dealing with measurements in science, different units are often used to express volume. It's essential to convert between these units properly to ensure calculations are accurate. In chemistry, you might frequently encounter milliliters (mL) and cubic centimeters (cm³), especially since these are common units for liquids. Thankfully, these two units are equivalent:
  • 1 mL = 1 cm³
This equivalence makes conversions straightforward. Simply remember, if you have a volume in milliliters, it's the same number in cubic centimeters.
For example, consider converting 125 mL of bromine to cm³. Since 1 mL equals 1 cm³, the calculation is simple:
  • 125 mL × \( \frac{1 \, \text{cm}^3}{1 \, \text{mL}} = 125 \, \text{cm}^3 \)
Whether you're dealing with liquids or gases, understanding how to convert volumes between units efficiently is a foundational skill in chemistry.
Mass Calculation
Calculating the mass of a substance involves using the concept of density, which is defined as mass per unit volume. The formula you need is:
  • Mass = Density × Volume
This formula is a fundamental relationship in chemistry that helps you find how much a substance weighs based on its density and volume.
Let's take bromine as an example. If you have a density of 3.12 g/cm³, and you want to find the mass of 125 cm³ of bromine:
  • Mass = \( 3.12 \frac{\text{g}}{\text{cm}^3} \times 125 \text{ cm}^3 = 390 \, \text{g} \)
The mass of 125 mL (or cm³) of bromine is 390 grams. Calculating mass using density and volume is straightforward as long as you ensure the units are compatible.
Chemical Elements
Understanding chemical elements, like bromine, involves recognizing their properties, such as density and state of matter at room temperature. Bromine, for example, is a unique element:
  • It appears as a reddish-brown liquid at room temperature.
  • The symbol for bromine is \( \mathrm{Br} \).
  • Its molecular form is \( \mathrm{Br}_{2} \), meaning it's diatomic.
Familiarizing yourself with the properties of elements is crucial when performing chemical calculations. For instance, knowing bromine has a density of 3.12 g/cm³ allows you to compute how much space a specific mass of bromine will occupy.
Consider how to determine the volume of 85.0 grams of bromine using its density:
  • Volume = \( \frac{85.0 \, \text{g}}{3.12 \, \frac{\text{g}}{\text{cm}^{3}}} \approx 27.2 \, \text{cm}^3 \)
This understanding bridges the gap between theoretical knowledge of chemical properties and practical applications in laboratory settings.

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Most popular questions from this chapter

Make molecular-level (microscopic) drawings for each of the following. a. Show the differences between a gaseous mixture that is a homogeneous mixture of two different compounds, and a gaseous mixture that is a homogeneous mixture of a compound and an element. b. Show the differences among a gaseous element, a liquid element, and a solid element

Round off each of the following numbers to the indicated number of significant digits, and write the answer in standard scientific notation. a. 0.00034159 to three digits b. \(103.351 \times 10^{2}\) to four digits c. 17.9915 to five digits d. \(3.365 \times 10^{5}\) to three digits

Apothecaries (druggists) use the following set of measures in the English system: $$\begin{aligned} 20 \text { grains ap } &=1 \text { scruple (exact) } \\ 3 \text { scruples } &=1 \text { dram ap (exact) } \\ 8 \operatorname{dram} a p &=1 \text { oz ap (exact) } \\ 1 \operatorname{dram} a p &=3.888 g \end{aligned}$$ a. Is an apothecary grain the same as a troy grain? (See Exercise 49.) b. 1 oz ap 5 ________ oz troy. c. An aspirin tablet contains \(5.00 \times 10^{2} \mathrm{mg}\) of active ingre- dient. What mass in grains ap of active ingredient does it contain? What mass in scruples? d. What is the mass of 1 scruple in grams?

Convert the following Kelvin temperatures to Celsius and Fahrenheit degrees. a. the temperature that registers the same value on both the Fahrenheit and Celsius scales, 233 K b. the boiling point of helium, 4 K c. the temperature at which many chemical quantities are determined, 298 K d. the melting point of tungsten, 3680 K

The density of an irregularly shaped object was determined as follows. The mass of the object was found to be \(28.90 \mathrm{g}=\) 0.03 \(\mathrm{g} .\) A graduated cylinder was partially filled with water. The reading of the level of the water was \(6.4 \mathrm{cm}^{3} \pm 0.1 \mathrm{cm}^{3}\) The object was dropped in the cylinder, and the level of the water rose to \(9.8 \mathrm{cm}^{3} \pm 0.1 \mathrm{cm}^{3} .\) What is the density of the object with appropriate error limits? (See Appendix \(1.5 . )\)

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