the top of a small table is 0.500 m 2. a.calculate how many square inches it is (1.000 inch =2.540...

The top of a small table is 0.500 m2.

a. Calculate how many square inches it is (1.000 inch =2.540 cm)

b.Is the table top a pure substance?

c. If the table is sawed in half, is that a physical or chemical change?

d.If the table is burned, is that a physical or chemical change?

e. If the table floats in water (before being burned), what does that tell you about the spacing of the molecules compared to water?

1. Is ice water homogeneous or heterogeneous?

2. Is freezing a chemical or physical change?

3. Is baking cookies a chemical or physical change?

4. How do molecules differ in a liquid than a solid?

5. A soda can has a radius of 3.25 cm and a height of 12.00 cm.

a. Calculate the volume of the can (398 cm3)

b. The can has a mass of 557 g. Calculate the density. (1.39 g/cm3)

c. Convert the density to kg/m3. (1390 kg/m3)

1. Pure Substances – either all one element or all one compound. Have definite unchanging properties

2. Types

a) Elements –Au, O2

b) Compounds – pure water

Pure Substances

1.Uniform mixture of two or more elements and compounds (only one phase)

2.Types

a) Solutions – liquid homogeneous mixtures

Examples Solute Solvent

Kool-Aid

Salt Water

Homogeneous Mixtures

b) Alloys – Solid uniform mixtures. Usually metals.

Stainless Steel – Iron and Chromium

Brass – Copper and Zinc

Bronze – Copper and tin

Homogeneous Mixtures

1. Nonuniform mixture, composed of two or more phases

2. Phase – One physical state with definite boundaries3. Examples:

Ice Water – Two Phases (water & ice)Italian dressing – More than two phasesGranite – Multiple phasesZinc and Sulfur

Heterogeneous Mixtures

1. Can be observed without changing the substance into another substance

2. Examples:Melting Point/Freezing PointBoiling PointHardnessSolubilityMalleabilityDuctility

Physical Properties

Position Volume Attraction

Solid Fixed (close together)

Definite Strong (close together)

Liquid Atoms wander

(close together)

Definite (takes shape of container)

Strong (close together)

Gas Atoms wander (far apart and fast)

Expand to fill container (can be compressed)

Weak or no attraction (too far apart)

Plasma Ions and electrons wander

Expand to fill container

Weak or no attraction (too far apart)

Solids

Can “wiggle” in place (these are the wiggle lines)

Liquids

They wander in random patterns quite close to one another.

Gases

Plasma• Plasma - 4th state of matter

• Ionized gases– Electrons removed from the atoms– Positive ions remain

• Present in:– Stars– Lightning– Arc welding

• Most common state of matter in the universe

hydrogen and helium plasma (sun)

ee

e

ee

eHe+

H+ H+

H+ H+

He+

He+

He+ e

e

Plasma

Can an elephant swim?

Can an elephant swim?

1. History – Archimedes story

2. Density = mass per unit volume

3. D = m/V

Mass Grams

V mL or (cm3)

D g/mL

Physical Properties : Density

4. Intensive property – does not depend on amount present

5. Volume formulas

Rectangular

Cylinder

Irregular shape (Archimedes)

6. Solving for variables

a. Algebra

3 = x/8 2 = 6/x

b. Density

Solve for m

Solve for V

1. A ring has a mass of 8.99 g and a volume of 0.796 mL. Is the ring gold (19.3 g/mL)? (Ans: 11.3 g/mL)

2. A substance masses 47.5 g. It is put into a grad cylinder containing 12.5 mL of water. After immersing the substance, the total volume is 31.8 mL. What is the density? (Ans: 2.46 g/mL)

3. Salt has a density of 2.16 g/mL. What is the volume of 485 g of salt? (Ans: 225 mL)

4. What is the mass of 1.520 L of kerosene ( = 0.8200 g/mL)? (Ans: 1246 g)

5. What volume would 0.450 kg of lead occupy ( = 11.3 g/cm3) (Ans: 39.8 cm3)

6. A cylinder has a radius of 2.00 cm and a height of 10.0 cm. If the cylinder is filled with 0.289 grams of a gas, density? (Ans: 0.00230 g/cm3)

A piece of wood that measures 3.0 cm by 6.0 cm by 4.0 cm has a mass of 80.0 grams.

a. Calculate the density of the wood.

b.Would the piece of wood float in water?

Calculate the mass of a cylinder of lead that is 2.50 cm in diameter, and 5.50 cm long. The density of lead is 11.4 g/mL.

A rectangular sample of aluminum has the measurements 1.34 cm by 2.58 cm by 10.00 cm.

a. Calculate the volume of the sample. (34.6 cm3)

b.Calculate the mass of the sample. The density of aluminum is 2.70 g/cm3. (93.3 g)

c. Would the block of aluminum float in mercury ( = 13.6 g/mL)?

d.An irregular sample of aluminum is found to have a mass of 50.00 grams. The sample is placed in 10.00 mL of water. Calculate the final volume reading of the graduated cylinder. (28.5 mL)

1. Chemical Change – one substance changed into another substance

2. Chemical Properties – tendency of a substance to react with other substances

3. ExamplesFlammabilityWill it rust (oxidize)?Acid or base?

Chemical Properties

4. Atoms rearrange and bonds are made and broken:

2H2 + O2 2H2 O

Recognizing a chemical change:

1. Properties change (rust is different than iron)

2. Gets hotter or colder• Exothermic – Gives off heat (burning gas, exercise)

• Endothermic – Absorbs heat (cooking)

3. Color change

4. Gas given off

5. Light produced (glow stick)

A. Antoine Lavosier (1789)

B. Law of cons. of mass – mass is neither created nor destroyed in a chemical reaction

Law of Conservation of Mass

C. Two Examples

1. 2H2 + O2 2H2O

10g + 80g 90g

2. Burning wood ?

Wood + O2 Gas + Ash.

Law of Conservation of Energy

Energy can never be created or destroyed in a chemical reaction. It can only change form.

1. Battery: turns chemical to electrical to mechanical energy.

2. ALWAYS lose some energy in transformations (usually as waste heat)

Law of cons. of mass/energy – mass and energy cannot be created or destroyed. They only change form

E = mc2

E= Energy

m = mass

c = speed of light

Einstein:1905

2. A small amount of matter can be destroyed to release a large amount of energy (nuclear processes) (20 g U 18 g U)

Energy – The capacity to do work– 1 Joule = 1 Newton-meter– 1 calorie = amount of heat to raise one gram of

water by 1o C

1 calorie = 4.18 Joule

(1 nutritional Calorie = 1000 calories, 1 kilocalorie)

Two types

• Potential Energy - Stored

Chemical Energy – energy stored in chemical bonds» Plants absorb energy from the sun» That energy is released through digestion/burning

• Kinetic Energy – energy in motion

Energy of moving atoms and molecules

Specific Heat – Amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius or Kelvin

• Unit – cal/goC or J/goC

• Symbol = Cp

• Higher the specific heat, the more energy needed to raise the temperature

Specific Heat

Predict the specific heat of the following (high or low):

Car hood

Pot of water

Beach sand

Plants & Trees

Substance Specific Heat (cal/goC)

Specific Heat (J/goC)

Water 1.000 4.184

Ice 0.492 2.06

Aluminum 0.214 0.895

Gold 0.031 0.129

Copper 0.092 0.385

Zinc 0.093 0.388

Iron 0.106 0.444

q = mCpT

q = heat in Joules

m = mass (grams)

Cp = specific heat (J/goC)

T = Tfinal – Tinitial

How much heat must be supplied to a 500.0 gram iron pan (Cp = 0.444 J/g oC) to raise its temperature from 20.0oC to 100.0oC?

q = mCpT

q = (500.0g)(0.444J/g oC)(100.0oC -20.0oC)

q = (500.0g)(0.444J/g oC)(80.0oC)

q = 17,800J or 17.8 kJ

Suppose we use a similar pan, except it is now made of Aluminum (Cp = 0.895 J/goC)?

q = mCpT

q = (500.0g)(0.895 J/goC)(100oC -20oC)

q = (500.0g)(0.895 J/goC)(80oC)

q = 35,800 J or 35.8 kJ

What temperature change would 50.0 g of rock undergo if they absorbed 452 Joules of heat? (Assume the specific heat is 0.836 J/goC).)

ANS: 10.8 oC (which is also 10.8 K)

A sample of copper (Cp = 0.092 cal/goC) undergoes a temperature change from 24.0oC to 76.0 oC upon the addition of 1300.0 Joules of heat.

a. Convert the heat to calories (311 cal)

b.Calculate the mass of the copper (65 g)

A 500.0 g sample of zinc absorbed 4850 J of heat. The temperature increased from 20.00 oC to 45.00 oC.

a. Calculate the specific heat of the sample in J/goC. (0.388 J/goC)

b. Calculate how many calories of heat the sample absorbed. (1160 cal)

c. A separate sample of zinc absorbs the same amount of heat. However, the temperature only rises from 20.00oC to 28.00oC. Calculate the mass of the sample in kilograms. (1.56 kg)

Heating Curves

1. Changes of state do not have a temperature change.

1. Melting/Freezing

2. Boiling/Condensing

2. A glass of soda with ice will stay at 0oC until all of the ice melts.

3. Graph “flattens out” during changes of state

Heat (Joules)

Temperature (oC) Boiling

Melting

Ice warms up

Water warms up

Steam heats up

1. Calculate the square root of 57.2 (7.56)

2. Calculate the cube root of 57.2 (3.85)

3. Calculate the cube root of 144 (5.24)

4. Calculate the cube root of 1,728 (12)

5. Calculate the 4th root of 625 (5)

6. Calculate the 4th root of 65,536 (16)

1. What is the volume of 0.0354 kg of gold? (1.83 cm3)

2. What mass of mercury would occupy 25.0 mL (340g)

3. Which would have a greater mass, 50 mL of water (1g/mL) or 50 mL of vegetable oil (0.72 g/mL)

4. A piece of zinc (7.14 g/cm3) has a mass of 257 g. If the zinc is rectangular and has a length of 1.00 cm and a width of 4.00 cm, what is the height? (9.00 cm)

5. A cylinder of copper (8.92 g/cm3) is found to have a mass of 1681 grams. The cylinder is 15.0 cm tall

a. Calculate the radius. (2.00 cm)

b. Would the sample float in water?

c. Would the sample float in mercury?

d. Suggest a method for measuring the volume of the cylinder in the lab.

A 250.0 g sample of water is heated from 21.20 oC to 98.30 oC to make tea.

a. Calculate the heat required in kJ. (80.6 kJ)

b. Calculate how many calories of heat the sample absorbed. (19.3 kcal)

c. When the same amount of heat is added to 600.0 grams of an unknown substance, the temperature rises by 59.7 oC. Calculate the specific heat of the unknown substance. (2.25 J/goC)

d. The unknown substance is rectangular and has dimensions 10.0 cm by 15.0 cm by 5.00 cm. Would the substance float in water? ( = 0.800 g/cm3)

5 a) Chemical

b) Physical

c) Physical

d) Chemical

e) Physical

7 a) Chemical

b) Chemical

c) Physical

d) Physical

8. Chemical (change in substance)

13. 7.91 g/mL

15. 1.74 g/ml, 575 mL

16. 21.0 mL

20. 29.0 kg (29,000 g)

25. 1.11 g/mL

26. 2.51 mL

29. Lead = 88.5 cm3, Gold = 51.8 cm3

65. 0.896 J/goC, Al

67. 2.05 J/goC

71. 1.4 oC

73. 46.7 g

74. 0.0317 cal/goC or 0.133 J/goC

83. 28.3 mL

90. 25.5 kcal

92. 0.242 cal/goC or 1.02 J/goC

63. 0.853 J/goC

64. 0.031 cal/goC

65. 0.896 J/goC, Al

66. 58 oC

67. 2.05 J/goC

68. 506 J

71. 1.4 oC

72. 9.38oC(Fe) 32oC(Au) 0.997o(H2O)

73. 46.7 g

92. 0.242 cal/goC or 1.02 J/goC

74. 0.0317 cal/goC or 0.133 J/goC

83. 28.3 mL

90. 25.5 kcal

19.1.00 gal = 3.785 L

3.785 L X 1000 mL = 3785 mL

1 L

D = M/V

M = DV = (0.67 g/mL) X 3785 mL = 2536 g

2536 g X 1 lb = 5.591 lb

453.6g

20.1.50 L = 1500 mL

M = DV = (19.3 g/mL)(1500 mL) = 29,000 g

29,000 g X (1 kg/1000 g) = 29.0 kg

1. 2.70 g/mL

2. 2.33 mL

3. 135 g

4. 4.20 g/mL

5. 3.92 g/mL

6. 80.0 g

7. 2.08 g/mL

8. 26.5 g

9. 7.20 g/mL

10. 2.24 g/mL

Chapter 3 Problem Sheet – Density

a)1.01 g/mL k) 1.96 g/L

b)158 g l) 11.3 g/mL

c)7.92 g/mL m) 0.0018 g/mL

d)2.70 g/mL n) 0.0022 cm

e)331 g

f) 2.16 g/mL

g)13.6 g/mL

h)225 g

i) 1.25 g/L

j) 1.59 g/mL

Measuring

Measuring

Measuring

“I would measure with a ruler with more marks to get more significant figures. ”

“More supplies would make the experiment go faster.”

“The metal pieces should be allowed to carefully slide down the side of the cylinder. This would prevent splashing of the water. The splashing seen in this experiment probably resulted in a smaller volume for the metal pieces.

“I would do more trials to get more numbers. This should give a more accurate average.”

Answers to Review Sheet:

1. (B) 6. (A)

2. (E) 7. (D)

3.(C) 8. (B)

4. (E) 9. (D)

5. (A) 10. (B)