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Gridlocks: 101 printable chemistry puzzles

  • 2 Puzzles for 11-14 year olds: from acids to states of matter
  • 3 Puzzles for 14-16 year olds: from atoms to units of volume
  • 4 Puzzles for 16-18 year olds: from aqueous ions to subshells

101 sudoku-style chemistry puzzles with printable worksheets and answers to engage your students and consolidate their knowledge about key topics

Gridlocks are a fun and stimulating way for students to learn the facts they need in chemistry. Discover 101 printable puzzles with answer sheets, covering core topics for ages 11–14, 14–16 and 16+.

Download the puzzles for each age group below, or read on to find out:

  • How gridlock puzzles work
  • How you can use the puzzles in your teaching
  • What your students can learn

Download the puzzles

Browse, print and download the puzzles for your students’ age group:

11–14 years | 14 –16 years | 16–18 years

Each download includes a series of puzzles focusing on a particular topic, with a printable student worksheet and answers.

How do gridlock puzzles work?

  • Students begin by filling in a table to review the key ideas they need to complete the puzzles. The table contains information about a group of objects, concepts or things related to the chosen topic. To complete the table, students identify items in this group or match them with further information, data or examples.
  • After filling in the table, students use this information to complete the gridlock puzzles that follow in the worksheet.

Completing the puzzles

  • Each puzzle features a 4 x 4 grid divided into rows, columns and four 2 x 2 boxes.
  • The objective is to fill in the grid using information from the table at the top of the worksheet, so that each row, column and 2 x 2 box contains only one reference to any single item (or row) from the table.
  • Each 2 x 2 box is labelled using headings from the information table. These headings tell students what type of information should be used in that box.
  • Each puzzle includes instructions telling students whether to use the whole information table, or only a part of it.

How can I use these puzzles in my teaching?

You can use gridlock puzzles during lessons or set them as homework. They are designed as follow up activities to consolidate students’ knowledge, rather than as introductions to a topic. Ideally, students should have met at least some of the data the gridlock puzzles are based on already.

The worksheets are simple to set and can readily be peer or self assessed. During lessons, the puzzles can be used flexibly as part of an individual, group or class-based activity. You can also add an extra element of competition or challenge. For example, set a target time and invite students to try to beat the clock, or encourage groups of students to see who can solve the most.

What will my students learn?

Topic knowledge.

Each puzzle focuses on a topic appropriate to 11–14, 14–16 or 16–18 year old students. To solve the puzzle, students need to engage with the factual information the gridlock is based on, recalling the relationships between ideas and data established in the first part of the activity. For example, they need to recall that three electron pairs corresponds to trigonal planar geometry, or that sulfuric acid forms sulfate salts.

As they work on the puzzles, students will find themselves referring to the initial data repeatedly, gradually consolidating their knowledge of the relevant facts.

Problem-solving and thinking skills

Gridlock puzzles give students a problem-solving context for their learning, promoting engagement and offering students a sense of satisfaction in completing the grid. The puzzles also develop some important thinking skills, as students must use logical reasoning to survey the data given in the gridlock and determine which squares can be filled in.

Additional information

This resource was originally part of the Gridlocks microsite, produced by the Royal Society of Chemistry with support from the Wolfson Foundation.

A collage of screenshots of completed gridlock puzzles

Puzzles for 11-14 year olds: from acids to states of matter

A screenshot showing a completed gridlock puzzle, with a grid containing the names and formulas of four alkanes

Puzzles for 14-16 year olds: from atoms to units of volume

A screenshot showing a completed gridlock puzzle, with a grid containing information about hybrid orbitals in molecular geometry

Puzzles for 16-18 year olds: from aqueous ions to subshells

  • 11-14 years
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  • 16-18 years
  • Elements and the periodic table
  • Equations, formulas and nomenclature

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  • Q 1 An organized ways of solving problems and it involves atleast 6 steps is? Scientific Method Law Theory Hyphothesis 30 s
  • Q 2 Which of the following s a mixture? Methanol 7 -up Silica Ethanol 30 s
  • Q 3 Which of these cannot be separated into two or more substances by physical means alone? Whiskey Cow's Milk Kerosene Table Salt 30 s
  • Q 4 An example of a chemical property is Reactivity Taste Odor Color 30 s
  • Q 5 What element is requried for synthesis of hemoglobin? Iodine Copper Calcium Chronium 30 s
  • Q 6 Which of these can be separated into two or more elements only by a chemical process? Gold Nitrogen Pure Water Hydrogen 30 s
  • Q 7 If the temperature of an object is 0 kelbin, on temperature in celcious scale is  212 32 100 273 30 s
  • Q 8 Which of the following contains a single type of atom? Baking Soda Aluminum Carbon Disulfide Ether 30 s
  • Q 9 Which statement describe a chemical change? Reactivity Flammability solubility Iron wool burns in air when it is placed in a flame 30 s
  • Q 10 Which of the following is a compound? Sugar Air Seawater Chocolate 30 s
  • Q 11 How many significant figures are in the number 0.0234? 6 4 7 3 30 s
  • Q 12 It is defined as capacity for doing work is called Energy Pressure Friction Force 30 s
  • Q 13 It is the energyextracted and electricity generation from natural steam or hot water refer to Coal Geothermal Seawater Natural gas 30 s
  • Q 14 Law of conservation of mass states that in an ordinary chemical reactions, the mass of the reactants equals the mass of the products. John Dalton Antoine Lavoisier Dmitri Mendeleev Berthollet & Proust 30 s
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Chemistry Problems

Use chemistry problems as a tool for mastering chemistry concepts. Some of these examples show using formulas while others include lists of examples.

Acids, Bases, and pH Chemistry Problems

Learn about acids and bases. See how to calculate pH, pOH, K a , K b , pK a , and pK b .

  • Practice calculating pH.
  • Get example pH, pK a , pK b , K a , and K b calculations.
  • Get examples of amphoterism.

Atomic Structure Problems

Learn about atomic mass, the Bohr model, and the part of the atom.

  • Practice identifying atomic number, mass number, and atomic mass.
  • Get examples showing ways to find atomic mass.
  • Use Avogadro’s number and find the mass of a single atom .
  • Review the Bohr model of the atom.
  • Find the number of valence electrons of an element’s atom.

Chemical Bonds

Learn how to use electronegativity to determine whether atoms form ionic or covalent bonds. See chemistry problems drawing Lewis structures.

  • Identify ionic and covalent bonds.
  • Learn about ionic compounds and get examples.
  • Practice identifying ionic compounds.
  • Get examples of binary compounds.
  • Learn about covalent compounds and their properties.
  • See how to assign oxidation numbers.
  • Practice drawing Lewis structures.
  • Practice calculating bond energy.

Chemical Equations

Practice writing and balancing chemical equations.

  • Learn the steps of balancing equations.
  • Practice balancing chemical equations (practice quiz).
  • Get examples finding theoretical yield.
  • Practice calculating percent yield.
  • Learn to recognize decomposition reactions.
  • Practice recognizing synthesis reactions.
  • Practice recognizing single replacement reactions.
  • Recognize double replacement reactions.
  • Find the mole ratio between chemical species in an equation.

Concentration and Solutions

Learn how to calculate concentration and explore chemistry problems that affect chemical concentration, including freezing point depression, boiling point elevation, and vapor pressure elevation.

  • Get example concentration calculations in several units.
  • Practice calculating normality (N).
  • Practice calculating molality (m).
  • Explore example molarity (M) calculations.
  • Get examples of colligative properties of solutions.
  • See the definition and examples of saturated solutions.
  • See the definition and examples of unsaturated solutions.
  • Get examples of miscible and immiscible liquids.

Error Calculations

Learn about the types of error and see worked chemistry example problems.

  • See how to calculate percent.
  • Practice absolute and relative error calculations.
  • See how to calculate percent error.
  • See how to find standard deviation.
  • Calculate mean, median, and mode.
  • Review the difference between accuracy and precision.

Equilibrium Chemistry Problems

Learn about Le Chatelier’s principle, reaction rates, and equilibrium.

  • Solve activation energy chemistry problems.
  • Review factors that affect reaction rate.
  • Practice calculating the van’t Hoff factor.

Practice chemistry problems using the gas laws, including Raoult’s law, Graham’s law, Boyle’s law, Charles’ law, and Dalton’s law of partial pressures.

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  • Solve Avogadro’s law problems.
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Some chemistry problems ask you identify examples of states of matter and types of mixtures. While there are any chemical formulas to know, it’s still nice to have lists of examples.

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  • See examples of intrinsic and extrinsic properties of matter.
  • Get the definition and examples of solids.
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  • Get examples of chemical properties of matter.
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Molecular Structure Chemistry Problems

See chemistry problems writing chemical formulas. See examples of monatomic and diatomic elements.

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  • See how to calculate molecular mass.
  • Get examples of the monatomic elements.
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  • Calculate the number of atoms and molecules in a drop of water.

Nomenclature

Practice chemistry problems naming ionic compounds, hydrocarbons, and covalent compounds.

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  • Learn hydrocarbon prefixes in organic chemistry.

Nuclear Chemistry

These chemistry problems involve isotopes, nuclear symbols, half-life, radioactive decay, fission, fusion.

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Periodic Table

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Physical Chemistry

Explore thermochemistry and physical chemistry, including enthalpy, entropy, heat of fusion, and heat of vaporization.

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Spectroscopy and Quantum Chemistry Problems

See chemistry problems involving the interaction between light and matter.

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Stoichiometry Chemistry Problems

Practice chemistry problems balancing formulas for mass and charge. Learn about reactants and products.

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Unit Conversions

There are some many examples of unit conversions that they have their own separate page!

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1.8: Solving Chemical Problems

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    ↵

  Learning Objectives

  • Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities.
  • Describe how to use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties.
  • Convert between the three main temperature units: Fahrenheit, Celsius, and Kelvin.

It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:

\[\mathrm{speed=\dfrac{distance}{time}} \nonumber \]

An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of

\[\mathrm{\dfrac{100\: m}{10\: s}=10\: m/s} \nonumber \]

Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:

\[\mathrm{time=\dfrac{distance}{speed}} \nonumber \]

The time can then be computed as:

\[\mathrm{\dfrac{25\: m}{10\: m/s}=2.5\: s} \nonumber \]

Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”

These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method ). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers . This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.

Conversion Factors and Dimensional Analysis

A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor . For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,

\[\mathrm{\dfrac{2.54\: cm}{1\: in.}\:(2.54\: cm=1\: in.)\: or\: 2.54\:\dfrac{cm}{in.}} \nonumber \]

Several other commonly used conversion factors are given in Table \(\PageIndex{1}\).

When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:

\[\mathrm{34\: \cancel{in.} \times \dfrac{2.54\: cm}{1\:\cancel{in.}}=86\: cm} \nonumber \]

Since this simple arithmetic involves quantities , the premise of dimensional analysis requires that we multiply both numbers and units . The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield

\[\mathrm{\dfrac{in.\times cm}{in.}}. \nonumber \]

Just as for numbers, a ratio of identical units is also numerically equal to one,

\[\mathrm{\dfrac{in.}{in.}=1} \nonumber \]

and the unit product thus simplifies to cm . (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.

Example \(\PageIndex{1}\): Using a Unit Conversion Factor

The mass of a competition Frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table \(\PageIndex{1}\)).

If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.

\[x\:\mathrm{oz=125\: g\times unit\: conversion\: factor}\nonumber \]

We write the unit conversion factor in its two forms:

\[\mathrm{\dfrac{1\: oz}{28.349\: g}\:and\:\dfrac{28.349\: g}{1\: oz}}\nonumber \]

The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.

\[\begin{align*} x\:\ce{oz}&=\mathrm{125\:\cancel{g}\times \dfrac{1\: oz}{28.349\:\cancel{g}}}\\ &=\mathrm{\left(\dfrac{125}{28.349}\right)\:oz}\\ &=\mathrm{4.41\: oz\: (three\: significant\: figures)} \end{align*} \nonumber \]

Exercise \(\PageIndex{1}\)

Convert a volume of 9.345 qt to liters.

Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.

Example \(\PageIndex{2}\): Computing Quantities from Measurement Results

What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.

Since \(\mathrm{density=\dfrac{mass}{volume}}\), we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A \(\times\) unit conversion factor. The necessary conversion factors are given in Table 1.7.1: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:

\[\mathrm{9.26\:\cancel{lb}\times \dfrac{453.59\: g}{1\:\cancel{lb}}=4.20\times 10^3\:g}\nonumber \]

We need to use two steps to convert volume from quarts to milliliters.

  • Convert quarts to liters.

\[\mathrm{4.00\:\cancel{qt}\times\dfrac{1\: L}{1.0567\:\cancel{qt}}=3.78\: L}\nonumber \]

  • Convert liters to milliliters.

\[\mathrm{3.78\:\cancel{L}\times\dfrac{1000\: mL}{1\:\cancel{L}}=3.78\times10^3\:mL}\nonumber \]

\[\mathrm{density=\dfrac{4.20\times10^3\:g}{3.78\times10^3\:mL}=1.11\: g/mL}\nonumber \]

Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:

\[\mathrm{\dfrac{9.26\:\cancel{lb}}{4.00\:\cancel{qt}}\times\dfrac{453.59\: g}{1\:\cancel{lb}}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\:\cancel{L}}{1000\: mL}=1.11\: g/mL}\nonumber \]

Exercise \(\PageIndex{2}\)

What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?

\(\mathrm{2.956\times10^{-2}\:L}\)

Example \(\PageIndex{3}\): Computing Quantities from Measurement Results

While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.

  • What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
  • If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?

(a) We first convert distance from kilometers to miles:

\[\mathrm{1250\: km\times\dfrac{0.62137\: mi}{1\: km}=777\: mi}\nonumber \]

and then convert volume from liters to gallons:

\[\mathrm{213\:\cancel{L}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\: gal}{4\:\cancel{qt}}=56.3\: gal}\nonumber \]

\[\mathrm{(average)\: mileage=\dfrac{777\: mi}{56.3\: gal}=13.8\: miles/gallon=13.8\: mpg}\nonumber \]

Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:

\[\mathrm{\dfrac{1250\:\cancel{km}}{213\:\cancel{L}}\times\dfrac{0.62137\: mi}{1\:\cancel{km}}\times\dfrac{1\:\cancel{L}}{1.0567\:\cancel{qt}}\times\dfrac{4\:\cancel{qt}}{1\: gal}=13.8\: mpg}\nonumber \]

(b) Using the previously calculated volume in gallons, we find:

\[\mathrm{56.3\: gal\times\dfrac{$3.80}{1\: gal}=$214}\nonumber \]

Exercise \(\PageIndex{3}\)

A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).

  • What (average) fuel economy, in miles per gallon, did the Prius get during this trip?
  • If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?

Conversion of Temperature Units

We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The mercury or alcohol in a common glass thermometer changes its volume as the temperature changes. Because the volume of the liquid changes more than the volume of the glass, we can see the liquid expand when it gets warmer and contract when it gets cooler.

To mark a scale on a thermometer, we need a set of reference values: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).

Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example:

\[\mathrm{length\: in\: feet=\left(\dfrac{1\: ft}{12\: in.}\right)\times length\: in\: inches} \nonumber \]

  • y = length in feet,
  • x = length in inches, and
  • the proportionality constant, m, is the conversion factor.

The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (\(y = mx + b\)). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points (\(b\)).

The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as \(x\) and the Fahrenheit temperature as \(y\), the slope, \(m\), is computed to be:

\[\begin{align*} m &=\dfrac{\Delta y}{\Delta x} \\[4pt] &= \mathrm{\dfrac{212\: ^\circ F - 32\: ^\circ F}{100\: ^\circ C-0\: ^\circ C}} \\[4pt] &= \mathrm{\dfrac{180\: ^\circ F}{100\: ^\circ C}} \\[4pt] &= \mathrm{\dfrac{9\: ^\circ F}{5\: ^\circ C} }\end{align*} \nonumber \]

The y-intercept of the equation, b , is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:

\[\begin{align*} b&=y-mx \\[4pt] &= \mathrm{32\:^\circ F-\dfrac{9\:^\circ F}{5\:^\circ C}\times0\:^\circ C} \\[4pt] &= \mathrm{32\:^\circ F} \end{align*} \nonumber \]

The equation relating the temperature scales is then:

\[\mathrm{\mathit{T}_{^\circ F}=\left(\dfrac{9\:^\circ F}{5\:^\circ C}\times \mathit{T}_{^\circ C}\right)+32\:^\circ C} \nonumber \]

An abbreviated form of this equation that omits the measurement units is:

\[\mathrm{\mathit{T}_{^\circ F}=\dfrac{9}{5}\times \mathit{T}_{^\circ C}+32} \nonumber \]

Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:

\[\mathrm{\mathit{T}_{^\circ C}=\dfrac{5}{9}(\mathit{T}_{^\circ F}+32)} \nonumber \]

As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. The early 19th-century discovery of the relationship between a gas's volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases).

The freezing temperature of water on this scale is 273.15 K and its boiling temperature 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of \(\mathrm{1\:\dfrac{K}{^\circ\:C}}\). Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:

\[T_{\ce K}=T_{\mathrm{^\circ C}}+273.15 \nonumber \]

\[T_\mathrm{^\circ C}=T_{\ce K}-273.15 \nonumber \]

The 273.15 in these equations has been determined experimentally, so it is not exact. Figure \(\PageIndex{1}\) shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale.

Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking.

Example \(\PageIndex{4}\): Conversion from Celsius

Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?

\[\mathrm{K= {^\circ C}+273.15=37.0+273.2=310.2\: K}\nonumber \]

\[\mathrm{^\circ F=\dfrac{9}{5}\:{^\circ C}+32.0=\left(\dfrac{9}{5}\times 37.0\right)+32.0=66.6+32.0=98.6\: ^\circ F}\nonumber \]

Exercise \(\PageIndex{4}\)

Convert 80.92 °C to K and °F.

354.07 K, 177.7 °F

Example \(\PageIndex{5}\): Conversion from Fahrenheit

Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?

\[\mathrm{^\circ C=\dfrac{5}{9}(^\circ F-32)=\dfrac{5}{9}(450-32)=\dfrac{5}{9}\times 418=232 ^\circ C\rightarrow set\: oven\: to\: 230 ^\circ C}\hspace{20px}\textrm{(two significant figures)}\nonumber \]

\[\mathrm{K={^\circ C}+273.15=230+273=503\: K\rightarrow 5.0\times 10^2\,K\hspace{20px}(two\: significant\: figures)}\nonumber \]

Exercise \(\PageIndex{5}\)

Convert 50 °F to °C and K.

10 °C, 280 K

Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.

Key Equations

  • \(T_\mathrm{^\circ C}=\dfrac{5}{9}\times T_\mathrm{^\circ F}-32\)
  • \(T_\mathrm{^\circ F}=\dfrac{9}{5}\times T_\mathrm{^\circ C}+32\)
  • \(T_\ce{K}={^\circ \ce C}+273.15\)
  • \(T_\mathrm{^\circ C}=\ce K-273.15\)

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10.5k plays, 5th -  8th  , states of matter, 9th -  12th  .

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Chemistry - Unit 1: MATTER

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17 questions

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When rain freezes and turns to ice or sleet, this is an example of a

physical property

chemical property

physical change

chemical change

Two substances are combined. Gas is released, and a tall, black column forms. This is an example of a

Which answer below is an example of two physical properties of matter?

weight and toxicity

mass and reactivity

mass and volume

mass and pH

When two substances are combined, a milky substance forms and falls to the bottom of the flask. What would be the most likely reason this happened?

One of the substances dissolved.

The substances chemically reacted with one another.

Some of the liquid evaporated.

One of the substances crystallized.

Conductivity is an example of a/an

intensive property

extensive property

dependent property

To compress means to be squeezed or pressed together. Why is a gas easily compressible, but solids are not?

Gas particles are much smaller.

Particles in a solid are not strongly attracted to one another.

Molecules in a solid are much larger.

Particles in a gas are much farther apart compared to a solid.

Which state of matter has an indefinite shape and a definite volume?

Which state of matter has an indefinite shape and an indefinite volume?

Which state of matter has a definite shape and a definite volume?

is made of a single substance.

is made of more than one substance.

always has a fixed composition.

is made of substances that you can always see at all times.

Students were studying different liquid samples. Which of these is most likely a pure substance?

Liquid A was heated and evaporated completely. Nothing was left behind in the beaker.

An acid was mixed with Liquid B, and bubbles formed.

Once Liquid C cooled, a precipitate formed.

The density of Liquid D is 2.3 g/mL.

If a salad dressing with water and vinegar looks the SAME throughout (has a consistent color and texture), it would be an example of a

heterogeneous mixture

pure substance

homogeneous mixture

What happens when you mix water and NaCl?

NaCl is salt. It will dissolve in water.

NaCl is salt, and it will all just sink to the bottom of the container.

NaCl is salt, and it will all float on top of the water.

We have no way of knowing what NaCl is.

When salt dissolves in water, it is an example of a

chemical reaction

TRUE or FALSE: A salt water solution will look the same throughout.

Mixtures are

physically combined and can usually be separated.

physically combined and can never be separated.

chemically combined.

We have no way of knowing how they combine.

A salt water solution can be separated by

a centrifuge

evaporation

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