Download presentation

1
**I. Using Measurements (p. 44 - 57)**

CH. 2 - MEASUREMENT I. Using Measurements (p ) C. Johannesson

2
**A. Accuracy vs. Precision**

Accuracy - how close a measurement is to the accepted value Precision - how close a series of measurements are to each other ACCURATE = CORRECT PRECISE = CONSISTENT C. Johannesson

3
**B. Percent Error Indicates accuracy of a measurement your value**

accepted value C. Johannesson

4
**B. Percent Error % error = 2.9 %**

A student determines the density of a substance to be 1.40 g/mL. Find the % error if the accepted value of the density is 1.36 g/mL. % error = 2.9 % C. Johannesson

5
**C. Significant Figures Indicate precision of a measurement.**

Recording Sig Figs Sig figs in a measurement include the known digits plus a final estimated digit 2.35 cm C. Johannesson

6
**C. Significant Figures Counting Sig Figs (Table 2-5, p.47)**

Count all numbers EXCEPT: Leading zeros Trailing zeros without a decimal point -- 2,500 C. Johannesson

7
**Counting Sig Fig Examples**

C. Significant Figures Counting Sig Fig Examples 4 sig figs 3 sig figs 3. 5,280 3. 5,280 3 sig figs 2 sig figs C. Johannesson

8
**C. Significant Figures (13.91g/cm3)(23.3cm3) = 324.103g 324 g**

Calculating with Sig Figs Multiply/Divide - The # with the fewest sig figs determines the # of sig figs in the answer. (13.91g/cm3)(23.3cm3) = g 4 SF 3 SF 3 SF 324 g C. Johannesson

9
**C. Significant Figures 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL**

Calculating with Sig Figs (con’t) Add/Subtract - The # with the lowest decimal value determines the place of the last sig fig in the answer. 3.75 mL mL 7.85 mL 3.75 mL mL 7.85 mL 224 g + 130 g 354 g 224 g + 130 g 354 g 7.9 mL 350 g C. Johannesson

10
**C. Significant Figures Calculating with Sig Figs (con’t)**

Exact Numbers do not limit the # of sig figs in the answer. Counting numbers: 12 students Exact conversions: 1 m = 100 cm “1” in any conversion: 1 in = 2.54 cm C. Johannesson

11
**C. Significant Figures Practice Problems 5. (15.30 g) ÷ (6.4 mL)**

4 SF 2 SF = g/mL 2.4 g/mL 2 SF g g 18.1 g 18.06 g C. Johannesson

12
**D. Scientific Notation 65,000 kg 6.5 × 104 kg**

Converting into Sci. Notation: Move decimal until there’s 1 digit to its left. Places moved = exponent. Large # (>1) positive exponent Small # (<1) negative exponent Only include sig figs. C. Johannesson

13
**D. Scientific Notation Practice Problems 7. 2,400,000 g 8. 0.00256 kg**

9. 7 10-5 km 104 mm 2.4 106 g 2.56 10-3 kg km 62,000 mm C. Johannesson

14
**D. Scientific Notation Calculating with Sci. Notation**

(5.44 × 107 g) ÷ (8.1 × 104 mol) = Type on your calculator: EXP EE EXP EE ENTER EXE 5.44 7 8.1 ÷ 4 = = 670 g/mol = 6.7 × 102 g/mol C. Johannesson

15
**E. Proportions Direct Proportion Inverse Proportion y x y x**

C. Johannesson

16
**II. Units of Measurement (p. 33 - 39)**

CH. 2 - MEASUREMENT II. Units of Measurement (p ) C. Johannesson

17
**A. Number vs. Quantity Quantity - number + unit UNITS MATTER!!**

C. Johannesson

18
**B. SI Units Quantity Symbol Base Unit Abbrev. Length l meter m Mass m**

kilogram kg Time t second s Temp T kelvin K Amount n mole mol C. Johannesson

19
**B. SI Units Prefix Symbol Factor mega- M 106 kilo- k 103 BASE UNIT ---**

100 deci- d 10-1 centi- c 10-2 milli- m 10-3 micro- 10-6 nano- n 10-9 pico- p 10-12 C. Johannesson

20
**M V D = C. Derived Units 1 cm3 = 1 mL 1 dm3 = 1 L**

Combination of base units. Volume (m3 or cm3) length length length 1 cm3 = 1 mL 1 dm3 = 1 L D = M V Density (kg/m3 or g/cm3) mass per volume C. Johannesson

21
D. Density Mass (g) Volume (cm3) C. Johannesson

22
**Problem-Solving Steps**

1. Analyze 2. Plan 3. Compute 4. Evaluate C. Johannesson

23
**D. Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = (13.6 g/cm3)(825cm3)**

An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass. GIVEN: V = 825 cm3 D = 13.6 g/cm3 M = ? WORK: M = DV M = (13.6 g/cm3)(825cm3) M = 11,200 g C. Johannesson

24
**D. Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL**

A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid? GIVEN: D = 0.87 g/mL V = ? M = 25 g WORK: V = M D V = g 0.87 g/mL V = 29 mL C. Johannesson

25
**III. Unit Conversions (p. 40 - 42)**

CH. 2 - MEASUREMENT III. Unit Conversions (p ) C. Johannesson

26
**A. SI Prefix Conversions**

1. Find the difference between the exponents of the two prefixes. 2. Move the decimal that many places. To the left or right? C. Johannesson

27
**A. SI Prefix Conversions**

= 532 m = _______ km 0.532 NUMBER UNIT NUMBER UNIT C. Johannesson

28
**A. SI Prefix Conversions**

Symbol Factor mega- M 106 kilo- k 103 BASE UNIT --- 100 deci- d 10-1 move left move right centi- c 10-2 milli- m 10-3 micro- 10-6 nano- n 10-9 pico- p 10-12 C. Johannesson

29
**A. SI Prefix Conversions**

0.2 1) 20 cm = ______________ m 2) L = ______________ mL 3) 45 m = ______________ nm 4) 805 dm = ______________ km 32 45,000 0.0805 C. Johannesson

30
**B. Dimensional Analysis**

The “Factor-Label” Method Units, or “labels” are canceled, or “factored” out C. Johannesson

31
**B. Dimensional Analysis**

Steps: 1. Identify starting & ending units. 2. Line up conversion factors so units cancel. 3. Multiply all top numbers & divide by each bottom number. 4. Check units & answer. C. Johannesson

32
**B. Dimensional Analysis**

Lining up conversion factors: = 1 1 in = 2.54 cm 2.54 cm cm 1 = 1 in = 2.54 cm 1 in in C. Johannesson

33
**1. Dimensional Analysis 1.00 qt 1 L 1.057 qt 1000 mL 1 L = 946 mL qt**

How many milliliters are in 1.00 quart of milk? qt mL 1.00 qt 1 L 1.057 qt 1000 mL 1 L = 946 mL C. Johannesson

34
**2. Dimensional Analysis 1.5 lb 1 kg 2.2 lb 1000 g 1 kg 1 cm3 19.3 g**

You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3. lb cm3 1.5 lb 1 kg 2.2 lb 1000 g 1 kg 1 cm3 19.3 g = 35 cm3 C. Johannesson

35
**3. Dimensional Analysis 75.0 in3 (2.54 cm)3 (1 in)3 1 L 1000 cm3**

How many liters of water would fill a container that measures 75.0 in3? in3 L 75.0 in3 (2.54 cm)3 (1 in)3 1 L 1000 cm3 = L C. Johannesson

36
**4. Dimensional Analysis 8.0 cm 1 in 2.54 cm = 3.1 in cm in**

4) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? cm in 8.0 cm 1 in 2.54 cm = 3.1 in C. Johannesson

37
**5. Dimensional Analysis 550 cm 1 in 2.54 cm 1 ft 12 in 1 yd 3 ft**

5) Taft football needs 550 cm for a 1st down. How many yards is this? cm yd 550 cm 1 in 2.54 cm 1 ft 12 in 1 yd 3 ft = 6.0 yd C. Johannesson

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google