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Unit 4: Chemical Calculations. Conversion Factors Problem solutions are given in this format: Problem solutions are given in this format: – A x B x D.

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Presentation on theme: "Unit 4: Chemical Calculations. Conversion Factors Problem solutions are given in this format: Problem solutions are given in this format: – A x B x D."— Presentation transcript:

1 Unit 4: Chemical Calculations

2 Conversion Factors Problem solutions are given in this format: Problem solutions are given in this format: – A x B x D = F C E C E To solve you enter it into the calculator as follows: To solve you enter it into the calculator as follows: – A x B ÷ C x D ÷ E = F General idea: every numerator value is multiplied and every denominator value is divided. General idea: every numerator value is multiplied and every denominator value is divided.

3 Let’s Practice! Solve practice problems 1-3 now Solve practice problems 1-3 now

4 Metric Calculations Common Metric Abbreviations Common Metric Abbreviations Metric UnitAbbreviation nanogramng microgramµg milligrammg gramg kilogramkg millilitermL literL meterm centimetercm millimetermm

5 Metric Calculations Common Metric Prefixes Common Metric Prefixes PrefixMeaningExample Conversion Value nano-billionthnanogram 1,000,000,000 ng = 1 g micro-millionthmicrogram 1,000,000 µg = 1 g milli-thousandthmilliliter 1,000 mL = 1 L centi-hundrethcentimeter 100 cm = 1 m kilo-thousandkilogram 1,000 grams = 1 kg

6 Let’s Practice! Solve practice problems 4-9 Solve practice problems 4-9

7 Metric Calculations Common Measurement Conversions Common Measurement ConversionsLength: 1 mile (mi)=5,280 feet (ft) 1 kilometer (km)=1,000 meters (m) 1 centimeter (cm)=10 millimeters (mm) 1 meter (m)= 39.38 inches (in) 1 inch(in)=2.54 centimeters (cm) Volume: 1 gallon (gal)=4 quarts (qt) 1 quart (qt)=0.95 liters (l) 1 quart (qt)=32 ounces (oz) Mass: 1 kilogram (kg)=2.2 pounds (lb) 454 grams (g)=1 pound (lb) 1 pound (lb)=16 ounces (oz)

8 Dimensional Analysis Use factor-label method Use factor-label method – Example 1: How many seconds are there in 2.3 days?

9 Dimensional Analysis – Example 2: Calculate the number of milliliters in 10.2 gallons of gasoline.

10 Let’s Practice! Solve practice problems 10-15 Solve practice problems 10-15

11 Scientific vs. Standard Notation Scientific Notation: Scientific Notation: 54=5.4 x 10 1 70,789=7.0789 x 10 4 0.0898=8.98 x 10 -2 Standard Notation: Standard Notation: 2.44 x 10 5 =244,000 7.053 x 10 -2 =0.07053 6.8 x 10 1 =68 2.0 x 10 0 =2.0

12 Let’s Practice! Solve practice problems 16-21 Solve practice problems 16-21

13 Calculator Practice Entering Scientific Notation into a calculator Entering Scientific Notation into a calculator – Example:

14 Let’s Practice! Solve practice problems 22-24 Solve practice problems 22-24

15 Significant Figures Digits in a final calculation that have physical meaning Digits in a final calculation that have physical meaning Digits from 1-9 are always significant Digits from 1-9 are always significant – 456 = 3 sig figs Zeros between two other significant digits are always significant Zeros between two other significant digits are always significant – 1009 = 4 sig figs One or more additional zeros to the right of both the decimal place and another significant digit are significant One or more additional zeros to the right of both the decimal place and another significant digit are significant – 100.0 = 4 sig figs – 100.00 = 5 sig figs Zeros used solely for spacing the decimal point (placeholders) are not significant Zeros used solely for spacing the decimal point (placeholders) are not significant – 100 = 1 sig fig – 5,500 = 2 sig figs

16 Let’s Practice! Solve practice problem 25 Solve practice problem 25

17 Calculating Significant Figures When x and ÷, keep only as many significant figures as there are the least of in the computation When x and ÷, keep only as many significant figures as there are the least of in the computation – 4.50 x 2.1 = 9.45  9.5 –1–1–1–1

18 Let’s Practice! Solve practice problem 26 Solve practice problem 26


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