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**Dimensional analysis: **A method to *convert* a given result from one system of units to another.

The video (by Tyler DeWitt) below is an introduction to dimensional analysis. It discusses the reasons why & how it can be used in every day life.

Exploring more of the video discussion, let's discuss another conversion factor. Here is an equivalent statement: 2.54 cm = 1 in

If you divide both sides of that equation by 2.54 cm, you get:

1 in

1 = ------------

2.54 cm

This expression is called a *unit factor*. Since 1 inch and 2.54 cm are exactly equivalent, multiplying any expression by this unit factor will not change its *value.*

*Problem:*

A pin has a length of 2.85 cm. What is the length in inches (in)?

1 in 2.85 in

2.85 ~~cm~~ X ------------ = ------------ = 1.12 in

2.54 ~~cm~~ 2.54 in

We start off with our given, which is 2.85 cm and just like in the video, **the centimeter units cancel out** to give inches for the result. If you notice, **the answer also has 3 significant figures** just like our given 2.85. **Remember the conversion factor are exact numbers** and would not be considered for significant figures due to having infinite numbers.

The next video (by Tyler DeWitt) goes more into detail in the explanation and instruction of how to do dimensional analysis.

Length |

1 meter = 1.0936 yards 1 centimeter = 0.39370 inch 1 inch = 2.54 centimeters (exactly) 1 kilometer = 0.62137 mile 1 mile = 5280 feet 1 mile = 1.6093 kilometers 1 angstrom = 10-10 meter 1 angstrom = 100 picometers |

Mass |

1 kilogram = 1000 grams 1 kilogram = 2.2046 pounds 1 pound = 453.59 grams 1 pound = 0.45359 kilograms 1 pound = 16 ounces 1 ton = 2000 pounds 1 ton = 907.185 kilograms 1 metric ton = 1000 kilograms 1 metric ton = 2204.6 pounds 1 atomic mass unit = 1.66056 x 10-27 kilograms |

Volume |

1 liter = 10-3 m3 1 liter = 1dm3 1 liter = 1.0567 quarts 1 gallon = 4 quarts 1 gallon = 8 pints 1 gallon = 3.7854 liters 1 quart = 32 fluid ounces 1 quart = 0.94633 liter |

Temperature |

0 K = –273.15 °C 0 K = –459.67 °F K = °C + 279.15 5 9 |

Energy |

1 joule = 1kg • m2/s2 1 joule = 0.23901 calorie 1 joule = 9.4781 x 10-4 btu (British thermal unit) 1 calorie = 4.184 joules 1 calorie = 3.965 x 10-3 btu 1 btu = 1055.06 joules 1 btu = 252.2 calories |

Pressure |

1 pascal = 1 N/m2 1 pascal = 1 kg/m • s2 1 atomsphere = 101.325 kilopascals 1 atmosphere = 760 torr (mm Hg) 1 atmosphere = 14.70 pounds per square inch 1 bar = 105 pascals |

Sometimes, dimensional analysis is not a one-step process and can take multiple steps to solve a single problem.

The video (by Tyler DeWitt) below discusses, explains, and instructs how to do multi-step unit conversion problems.

Let's work out a multiple conversion problem going step-by-step!

**A student has entered a 10.0 km run. How long is the run in miles?**

First, we know that we have to convert kilometers to miles, and then we know that our given is 10.0 km.

Next, we need to know our equivalence statements, which are:

1 km = 1000 m

1 m = 1.094 yd

1760 yd = 1 mi

Now that we know our equivalence statements, let's figure out the strategy. We can follow this process:

kilometers → meters → yards → miles

Alright! Let's try to follow this process going step by step:

*Kilometers to Meters*

1000 m

10.0 ~~km~~ X ------------ = 1.00 x 10^{4} m **Note the km unit cancel out to give the meters result.**

1 ~~km~~

*Meters to Yards*

1.094 yd Note the m unit cancel out to give the yd result. Also, there should be

1.00 x 10^{4} ~~m~~ X ------------ = 1.094 x 10^{4} yd **3 significant figures in the answer, but this is an intermediate step and the **** **

1 ~~m~~ FINAL result should round off to 3 significant figures.

*Yards to Miles*

1 mi **Note in this case that 1 mi equals exactly 1760 yd by designation, and therefore,**

1.094 x 10

1760

Since the given is 10.0 km, the result can only have 3 significant figures and should be rounded to 6.22 mi making the final answer:

**10.0 km = 6.22 mi**

Now if we combine the steps, it should look like this:

1000 ~~m~~ 1.094 ~~yd~~ 1 mi

10.0 ~~km~~ X ------------ X ------------ X ------------ = **6.22 mi**

1 ~~km~~ 1 ~~m~~ 1760 ~~yd~~

Helpful Tip:

In doing chemistry problems, you should __ ALWAYS__ include the units for the quantities used.

How do you do dimensional analysis problems with numbers that have a top and a bottom?

The video (by Tyler DeWitt) below will help explain and instruct on how to work out problems with numbers that have both a top and bottom unit.

Let's try out a problem of our own!

A car is advertised as having a gas mileage of 15 km/L. Convert to miles per gallon.

We already know how to do the initial part from changing km to mi, so let's start with doing that:

First, we need our equivalence statements:

1 km = 1000 m

1 m = 1.094 yd

1760 yd = 1 mi

1 L = 1.06 qt

4 qt = 1 gal

Now, let's do part one of our conversion problem:

15 ~~km~~ 1000 ~~m~~ 1.094 ~~yd~~ 1 mi **Note that we were able to cancel out the km unit to mi, but we still **

------------ X ------------ X ------------ X ------------ = 9.3238636 mi/L have yet to **convert L to gallon.**

L 1 ~~km~~ 1 ~~m~~ 1760 ~~yd~~

We had to make sure that we canceled out the units by putting the same unit on the bottom of the next part of the equation (e.g., 15 km is on the top of the first step and 1 km is on the bottom of the second, 1000 m is on the top of the second step and 1 m is on the bottom of the third step).

Now, we are going to reverse the process. We have 1 L on the bottom of the first step so we need to make sure that the next L unit is on the top of the next step.

9.3238636 mi 1 ~~L~~ 4 ~~qt~~ Remember that we have to round the final result to obtain the

--------------------- X ------------ X ------------ = 35.18439 mi/gal = **35 mi/gal correct number of significant figures which should be 2 for **

~~L~~ 1.06 ~~qt~~ 1 gal **this problem because our given is 15.**

Here is what it would look like if it is done all together:

15 ~~km~~ 1000 ~~m~~ 1.094 ~~yd~~ 1 mi 1 ~~L~~ 4 ~~qt~~

------------ X ------------ X ------------ X ------------ X ------------ X ------------ = 35.18439 mi/gal = 35 mi/gal

~~L~~ 1 ~~km~~ 1 ~~m~~ 1760 ~~yd~~ 1.06 ~~qt~~ 1 gal

Everything that we did in the first step is done in purple while everything that we did in the second step is done in green.

Again, it helps when you write down your units and are able to cross them out as you go. This helps check your process to make sure that you are doing it correctly. **ALWAYS INCLUDE UNITS!**

The answers are located in the boxes: Highlight the text inside the box to reveal answer

Here are some conversion factors that you should probably know:

1 km = 1000 m | 1 cm = 10 mm | 1 L = 1000 mL | 1 g = 1000 mg |

1 m = 1000 mm | 1 m = 100 cm | 1 kg = 1000 g |

Here are other conversion factors:

1 mi = 1.61 km | 1 L = 1.06 qt | 1 oz = 28.3 g | 1 mL = 20 drops | 1 cm3 = 1 mL |

1 qt = 0.946 L | 1 lb = 454 g | 1 ton = 2000 lbs | 1 gal = 4 quarts | 1 in = 2.54 cm |

1 kg = 2.2 lbs | 1 mi = 5280 ft | 365 days = 1 year |

Convert 25.0 g to kg.

1 kg 1000 g The answer should contain 3 significant figures because of the given 25.0. The leading zeros in the answer |

How many minutes are there in 1.6 years?

365 days 24 hr 60 min Because our given has 2 significant figures, our answer needs to have 2 as well. In this case, we can write the answer two different ways. The numbers that are underlined are the only numbers that count as significant figures. The trailing zeros in the first answer do not count because there is no decimal point in the result. |

If you are going 55 mph, what is your speed in m/s?

55 mi 1.61 km 1000 m 1 hr 1 min ------------ X ------------ X ------------ X ------------ X ------------ = 25 m/s1 hr 1 mi 1 km 60 min 60 s |

These questions were obtained from Hudson High School Honors Course worksheet.

There are three systems of measurement widely used for temperature:

- Celsius scale - used in physical sciences
- Kelvin scale - used in physical sciences
- Fahrenheit scale - used in many engineering sciences

*Note: The size of the temperature unit (degree) is the same for Kelvin and Celsius scales. The difference is the zero points.*

**Conversions:**

Temperature (Kelvin) = temperature (Celsius) + 273.15

or

Temperature (Celsius) = temperature (Kelvin) - 273.15

*Example:*

Convert 400.0 K to the Celsius scale.

400.00 K - 273.15 = 126.85 °C

Helpful Tip:

Kelvin does not use a degree symbol for its unit unlike Celsius and Fahrenheit; it is symbolized by the letter K.

The degree size and zero points are different when considering Fahrenheit. As seen in the illustration above, The degree size for Fahrenheit is 180° and 100° for Celsius. So, the unit factor is:

180 °F 9 °F

------- or -------

100 °C 5 °C

This is because 180 ÷ 100 = 1.8 and in fraction form that is 9/5. You will also have to use the reciprocal depending on the direction that is needed.

Now, for the difference in zero points. To obtain the temperature in Celsius, you must subtract 32 °F from the given temperature in Fahrenheit and then multiply the unit factor to adjust for the difference in degree size.

5 °C

(*T*_{F} - 32 °F) X ------- = *T*_{C}

9 °F

As you can see in this equation, the reciprocal was used.

If the degree is needed in Fahrenheit and you have Celsius, then the following equation is used.

9 °F

*T*_{F} = *T*_{C} X ------- + 32 °F

5 °C

*Example:*

Convert 98.6 °F to the Celsius and Kelvin scales.

First, convert to Celsius:

5 °C

(98.6 °F - 32 °F) X ------- = 37.0 °C

9 °F

Next, convert the Celsius answer you just obtained to the Kelvin scale:

*T*_{K} = *T*_{C} + 273.15

*T*_{K} = 37.0 °C + 273.15

*T*_{K} = 310.2 K

**Density: **A property of matter used as an "identification tag" for a substance - the mass of the substance per unit volume of the substance.

mass

Density = -------------

volume

Compound |
Density in g/cm^{3} at 20 °C |
---|---|

Chloroform | 1.492 |

Diethyl ether | 0.714 |

Ethanol | 0.789 |

Isopropyl alcohol | 0.785 |

Toluene | 0.867 |

*Example:*

Identify the unknown liquid. 25.00 cm^{3} of the substance has a mass of 19.625 g at 20 °C. Use the chart on the right to solve for what the unknown liquid could possibly be.

The first thing needed, is to find the density of the unknown liquid:

mass 19.625 g

Density = ------------- = ----------------------- = 0.7850 g/cm^{3}

volume 25.00 cm^{3 }

The density obtained matches with the density of isopropyl alcohol.

*Note: Ethanol is also very close and you would need to run more tests in order to ensure that isopropyl alcohol is the correct unknown liquid.*

Substance |
Physical State |
Density (g/cm^{3}) |

Oxygen | Gas | 0.00133 |

Hydrogen | Gas | 0.000084 |

Ethanol | Liquid | 0.789 |

Benzene | Liquid | 0.880 |

Water | Liquid | 0.9982 |

Magnesium | Solid | 1.74 |

Salt (sodium chloride) | Solid | 2.16 |

Aluminum | Solid | 2.70 |

Iron | Solid | 7.87 |

Copper | Solid | 8.96 |

Silver | Solid | 10.5 |

Lead | Solid | 11.34 |

Mercury | Liquid | 13.6 |

Gold | Solid | 19.32 |

FUN FACTS:

The liquid in your car's lead storage battery changes density when the sulfuric acid is consumed as the battery discharges. If the battery falls below a certain amount, the battery will have to recharge.

You can determine the amount of antifreeze, which tells you the level of protection against freezing in the cooling system of your car.

The center of a black hole (the singularity) is infinitely dense.

Saturn has the lowest density of all the planets in our solar system. Saturn has a density of 0.687 g/cm^{3} which is less than water. So, if you have a large enough pool filled with water, Saturn will float!

Silver has a density of 10.5 g/cm^{3} and gold has a density of 19.3 g/cm^{3}. Which would have a greater mass, 5 cm^{3} of silver, or 5 cm^{3} of gold?

Silver: mass mass = density X volume Gold: mass = (19.3 g/cm Answer: gold has more density. |

One of the body's responses to an infection or injury is to elevate its temperature. A certain flu victim has a body temperature of 102 °F. What is the temperature on the Celsius scale?

Converting °F to °C 5 °C 5 °C Answer: 38.9 °C |

An irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 2.0 mL. The height of the water rose to 7.0 mL. If the mass of the stone was 25 g, what was its density?

You must find the difference in the volume of water from the initial point to after the stone was dropped. 25 g Answer: density = 5 g/mL |

Convert -196 °C to the Kelvin scale.

Temperature (Kelvin) = temperature (Celsius) + 273.15 Answer: |