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Chapter 16: Problem 8

State the physical quantity corresponding to each of the following metric units. (a) \(\mathrm{km} / \mathrm{h}\) (b) \(\mathrm{kg} / \mathrm{m}^{3}\) (c) \(J /(k g \times K)\) (d) mol solute/1-L solution

Short Answer

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(a) Speed, (b) Density, (c) Specific Heat Capacity, (d) Molarity

Step by step solution

01

Analyze Unit for (a)

The unit given is \( \mathrm{km/h} \). This indicates kilometers per hour, which is a measure of speed or velocity. Speed is defined as the distance traveled per unit of time.
02

Analyze Unit for (b)

The unit given is \( \mathrm{kg/m^3} \). This indicates kilograms per cubic meter, which is a measure of density. Density is defined as the mass per unit volume.
03

Analyze Unit for (c)

The unit given is \( J/(kg \times K) \). This indicates joules per kilogram per Kelvin, which is a measure of specific heat capacity. Specific heat capacity is defined as the amount of heat needed to change the temperature of a unit mass of a substance by one degree Kelvin.
04

Analyze Unit for (d)

The unit given is mol solute/1-L solution. This refers to mole per liter, which is a measure of concentration. Specifically, it is known as molarity, which indicates the number of moles of solute per liter of solution.

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Key Concepts

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

Speed
Speed, often represented in units such as kilometers per hour (\(\mathrm{km/h}\)), is a fundamental concept of motion. It describes how fast an object travels over a specific distance. In scientific terms, speed is calculated as the distance an object moves divided by the time it takes to move that distance.
  • Formula: Speed = Distance/Time
  • Example: A car that covers 100 kilometers in 2 hours has a speed of 50 km/h.
Speed is a scalar quantity, meaning it only deals with magnitude (how fast), but not direction. This makes it different from velocity, which is a vector quantity and includes both speed and direction. For instance, saying a car is moving at 60 km/h tells us the speed, but saying it is moving at 60 km/h north gives us the velocity.
Density
Density is a measure of how much mass is contained in a given volume, typically expressed in kilograms per cubic meter (\(\mathrm{kg/m^3}\)). Understanding density involves not only how heavy an object is but how much space it occupies.
  • Formula: Density = Mass/Volume
  • Example: If a block of wood has a mass of 200 kg and occupies a space of 0.5 \(\mathrm{m^3}\), its density is 400 \(\mathrm{kg/m^3}\).
Density plays a crucial role in determining whether an object will float or sink when placed in a fluid. Objects with densities less than the fluid will float, while those with greater densities will sink. This principle is essential in fields ranging from material science to fluid dynamics.
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is required to raise the temperature of one kilogram of a substance by one degree Kelvin. It is expressed in joules per kilogram per Kelvin (\(J/(kg \times K)\)).
  • Formula: \( q = mc\Delta T \) (where \( q \) = heat energy, \( m \) = mass, \( c \) = specific heat capacity, and \( \Delta T \) = change in temperature)
  • Example: Water has a high specific heat capacity of approximately 4184 \(J/(kg \times K)\), meaning it takes a lot of energy to increase its temperature.
Specific heat capacity is a critical factor in climate science, engineering, and culinary arts, as it determines how substances respond to heat. A higher specific heat capacity means a material can absorb more heat without experiencing a significant increase in temperature.
Concentration
Concentration refers to the abundance of a constituent (usually a solute) within a mixture or solution. Molarity, a common unit of concentration, is measured in moles of solute per liter of solution (mol/L).
  • Formula: Molarity (M) = Moles of solute / Liters of solution
  • Example: If you dissolve 1 mole of sugar in 1 liter of water, the concentration (molarity) is 1 M.
Understanding concentration is crucial in chemistry, especially when performing reactions that require specific solute-to-solvent ratios. It helps chemists understand how reactions occur at the molecular level and predict the outcomes of chemical interactions. In industries, controlling concentration is vital for consistency in product manufacturing.

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Most popular questions from this chapter

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), is prepared industrially from the reaction of oxygen and isopropyl alcohol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\). Hydrogen peroxide reacts explosively with hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) and is used as a rocket fuel. The reactions for manufacturing hydrazine and the rocket explosion are as follows. $$\begin{array}{l}\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(l)+\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l) \\\2 \mathrm{H}_{2} \mathrm{O}_{2}(l)+\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\end{array}$$ Starting with \(50.0 \mathrm{~mL}\) of isopropyl alcohol \((d=0.786 \mathrm{~g} / \mathrm{mL})\) and excess other reactants, calculate: (a) the mass of nitrogen gas produced (b) the volume of nitrogen gas produced at STP (c) the mass of water produced

Draw a concept map that relates each of the following stoichiometric quantities: (a) volume of gaseous reactant and volume of gaseous product (b) volume of gaseous product and volume of gaseous reactant

Given that \(0.0755 \mathrm{~mol}\) of ammonia gas dissolves in \(0.155 \mathrm{~L}\) of solution, draw a mole concept map and calculate each of the following: (a) grams of \(\mathrm{NH}_{3}\) gas (at STP) dissolved in the solution (b) liters of \(\mathrm{NH}_{3}\) gas (at STP) dissolved in the solution (c) molecules of \(\mathrm{NH}_{3}\) gas dissolved in the solution (d) molar concentration of the ammonia solution

Calculate the molar mass of an unknown gas if an experiment yielded a value of \(1.45 \mathrm{~g} / \mathrm{L}\) at \(100{ }^{\circ} \mathrm{C}\) and \(0.989 \mathrm{~atm} .\)

Estimate an approximate answer for each of the following calculations. Verify your ballpark answer using a calculator: $$\begin{array}{l}\text { (a) } 0.750 \mathrm{~g} \mathrm{Mg} \times \frac{1 \mathrm{~mol} \mathrm{Mg}}{24.31 \mathrm{~g} \mathrm{Mg}}\times\frac{1\mathrm{~mol} \mathrm{MgO}}{1 \mathrm{~mol} \mathrm{Mg}} \times \frac{40.31 \mathrm{~g} \mathrm{MgO}}{1 \mathrm{~mol} \mathrm{MgO}} \\\=? \mathrm{~g} \mathrm{MgO}\end{array}$$
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