01C_Conversion Factors.pdf

(1)

Measurements and Calculations

(2)

(3)

Scientific Notation

 _{Used for very small}

& very big numbers

 The radius of a

carbon atom is 9.1 x 10-11_m

 The number of

oxygen atoms in 1 mL of water is 3.3 x 1022 atoms

9.1 x 10-11

is 0.000000000091 move the decimal 11 places to the right to put it after the 9

3.3 x 1022 _is

(4)

Big Numbers have Positive

Exponents

4.5 x 109 is 4,500,000,000

(5)

Big Numbers have Positive

Exponents

4.5 x 109 is 4,500,000,000

The exponent is equal to the number of places the decimal has been moved.

1

(6)

Small Numbers have Negative

Exponents

4.5 x 10-9 is 0.0000000045

The decimal point is moved to the position behind the

(7)

Small Numbers have Negative

Exponents

4.5 x 10-9 is 0.0000000045

The exponent is equal to the number of places the decimal has been moved with a - sign.

(8)

Using Scientific Notation in your

Calculator

 _{Find your EE or EXP}

button.

 Input the number 6.3 x 108 by putting in 6.3 then instead of x10 push the EE or EXP button

(9)

Using Scientific Notation in your

Calculator

 _{For a negative}

exponent use the +/-or (-) key, not the

subtraction key

 The read out on

your calculator may

(10)

Practice Problems

 Write the following in scientific notation:

– 0.0065 – 435,000 – 435.6

 Perform the following calculations and express the answer in scientific notation (remember to write the correct number of sig figs)

– (1.3 x 103_{)(5.724 x 10}4₎

– 4.503x103 _{+ 3.49x10}1 _{+ 5.5x10}2

(11)

Conversion Factors

 _{1.00 inch = 2.54 cm}  1.00 meter = 39.4 in

 _{1.00 km = 0.621}

mile

(12)

Problem Solving using

Conversion Factors

56.4 cm x --- = 22.2 inch1.00 in 2.54 cm

actual

(13)

Problem Solving using

Conversion Factors

56.4 cm x --- = 22.2 inch1.00 in 2.54 cm

cm unit

to start place cm unit

(14)

Problem Solving using

Conversion Factors

56.4 cm x --- = 22.2 inch1.00 in 2.54 cm

(15)

Density of a Liquid

 Determine the mass of the empty

cylinder (example: 45.321 g).

 Place a volume of liquid in a graduated cylinder. (50. mL)

 Determine the mass of the cylinder with 50. mL of the liquid ( example 94.372 g).

(19)

Density of a Liquid

 Mass of the liquid = mass of cylinder

with the liquid -mass of empty

cylinder = 94.372 g -45.321 g = 49.051 g

 _{Volume of liquid =}

50. mL

 _{Density of liquid =}

49.051g/50. mL = 0.98 g/mL

(20)

What is the density of the liquid?

A word problem

A graduated cylinder weighs

(21)

What is the density of the liquid?

A word problem

Volume of liquid = 10.0 mL

Mass of liquid = 39.359 g - 28.503 g mass = 10.856 g

(22)

How many significant figures?

a. 0.005020 b. 3.509

c. 2.6 x 10-3

(23)

Density of a Solid Rectangular

Block

 _{Measure mass on}

the balance. Mass = 78.347 g

 Measure

dimensions using a ruler. Volume =

length x width x height = 3.5 cm x 2.5 cm x 6.8 cm = 60. cm3

3.5 cm

2.5 cm 6.8 cm

(24)

Density of an Irregular Shaped

Solid (Metal)

 Place a volume of liquid in a graduated cylinder. (30. mL)

 Determine the mass of the metal (i. e.

104.372 g).

 Place the metal in the cylinder and note the volume of liquid and metal.

(25)

Density of an Irregular Shaped

Solid (Metal)

 Volume of liquid and metal = 44 mL

 Volume of metal = 44 mL - 30. mL = 14 mL

 D = 104.372 g/14 mL D = 7.5 g/mL

(26)

What is the density of the solid?

A word problem

A graduated cylinder has

20. mL of liquid in it. 41.802 g of metal is put in the cylinder which now reads 35 mL .

(27)

A word problem

Mass of metal = 41.802 g

Volume of metal = 35 mL – 20. mL Volume of metal = 15 mL

(28)

A word problem

Density = 41.802 g/15 mL = 2.8 g/mL

Could this be Al which has a density of 2.70 g/mL? YES

(29)

Calculations Using Density

Mass = Density x Volume

Density and volume are given in the problem. It may be necessary to

convert volume units to mL first.

EXAMPLE: How many grams does 2.5

(30)

Calculations Using Density

EXAMPLE: How many grams does 2.5 L of zinc weigh? (The density of zinc is 7.13 g/mL)

First convert; 2.5 L x 1000 mL/1 L = 2500 mL

Mass = 7.13 g/mL x 2500 mL = 18000 g

(31)

Calculations Using Density

Volume = Mass/Density

Density and mass are given in the problem. It may be necessary to convert mass units to grams first.

EXAMPLE: How many mL does 8.65 g

(32)

Calculations Using Density

EXAMPLE: How many mL does 8.65 g of zinc occupy? (The density of zinc is 7.13 g/mL)

No conversion is needed, the mass is in grams.

Volume = 8.65 g / 7.13 g/mL = 1.21 mL

(33)

Dosage Calculations

 _{Patients weight is}

given in pounds.

 Dosage is the number of mg of

medicine per pound of body weight

(34)

Dosage for a Solid Medication

A patient weighs 134 pounds. The dosage is 5.2 mg medicine/kg body weight. How many mg of medicine should be given to the patient?

(35)

Dosage for a Solid Medication

134 pounds x --- = 60.9 kg 1.00 kg

2.20 pound

60.9 kg x --- = 320 mg 5.2 mg med

1 kg patient

patient weight medicine

(36)

If the Solid Medication is in Pill

Form……….

The medication is given in 80. mg tablets means 1 tablet = 80. Mg

So for 320 mg medicine…..

320 mg med x --- = 4 tablets 1 tablet

(37)

In-class problems

What is the volume of 454 g of gold, which has a density of 19.3 g/mL?

(38)

Dosage for a Medication in

Solution

 _{A patient weigh 84.5}

pounds. The dosage is 6.50 mg medicine/ kg body weight. The medicine is in a

solution with 50.0

mg medicine per mL solution. How many mL should the

(39)

Dosage for a Medication in

Solution

Convert 84.5 pounds to 38.4 kg

Multiply 38.4 kg by the dosage to get 250. mg of medicine

(40)

Dosage for Medication in

Solution

84.5 lb x --- = 38.4 kg1.00 kg 2.20 lb

38.4 kg x --- = 250. mg6.50 mg 1 kg

250. mg x --- = 5.00 mL 50.0 mg

(41)

Temperature and Heat

 _{Temperature is a}

measure of how much motion the molecules have.

 Heat is the amount of energy required to change the

(42)

(43)

Temperature Conversions

Converting between Fahrenheit and Celcius

0_{F = (9/5)}0_{C + 32}

0_{C = (5/9)(}0F – 32)

Converting between Celcius and Kelvin

(44)

Temperature Practice

Liquid nitrogen has a temperature of about 77K. What is this temperature in Fahrenheit?

(45)

K = 0_{C + 273.15} 0_{F = (9/5)}0_{C + 32}

0_{C = K - 273.15} 0_{F = (9/5)(-196.15) + 32}

0_{C = 77 - 273.15} 0_{F = -353.07+ 32}

0_{C = -196.15} 0_{F = -353.07+ 32}

0_{F = -321.07}

-321 0F

(46)

Room Temperature is about 72 0F.

What is this temperature in Celcius? Kelvin?

0_{C = (5/9)(}0F – 32)

0_{C = (5/9)(72-32)}

0_{C = (5/9)(40.)}

0_{C = 22.222}