1st 6 Weeks Review

Test Wednesday

Remember your pencil and calculator

Part I.  Classify each of the following substances as; an element, a compound, a

solution, or a heterogeneous mixture.

Sand

Mixture

Salt

Compound

Pure Water

Compound

Soil

Mixture

Soda

Solution

Pure Air

Mixture

Carbon Dioxide

Compound

Gold

Element

Bronze

mixture

Oxygen

Element

Salad Dressing

Mixture

Salt Water

Solution

Part II.   In the spaces provided, describe the distinguishing characteristics of the

major categories of matter.• Element – A substance that cannot be

separated or broken down into simpler substances by chemical means; can be found on periodic table.

• Compound – A substance made up of atoms of two or more different elements joined by chemical bonds

• Solution – a homogenous mixture throughout which two or more substances are uniform.

• Mixture – a combination of two or more substances that are not chemically combined.

Part III.  Classification - State whether each of the following changes would be physical

or chemical.

Melting Ice

Physical

Burning Wood

Chemical

Breaking Glass

Physical

Painting Wood

Physical

Cooking

Chemical

Burning Propane

Chemical

Part IV.  Define - Use your own words to define each of the following terms in the

space provided.• Chemical Change – a change that occurs

when one or more substances change into entirely new substances with different properties.

• Physical Change – A change of matter from one form to another without a change in chemical properties.

• Change of State of Phase – When a substance changes from one state of matter to another.

Part V

1. 18 m = 1800 cm2. 167 mm = (typo change to m) 0.167 m3. 500 kg = 500,000 g4. 23 dm = 0.023 hm5. 1,589 dl = (typo change to kl) 0.1589 kl6. 700 ml = 0.000700 kl7. 5 cm = 50 mm8. 35.45 mg = 0.0003545 hg9. 0.5 l = 0.0005 kl10. 130 Dg = 13.0 kg

Density (#11)a) m = 453 g; V = 225 cm3

d = m/v = 453 g/225 cm3 = 2.01 g/ cm3

b) m = 5.0 g; V = 10.0 cm3

d = m/v = 5.0g/10.0 cm3 = 0.50 g/ cm3

c) m = 26.1 g; V = 2.0 mL

d = m/v = 26.1g/2.0 mL = 13.1 g/mL

#12 m = 75.2 g; V = 89.2 mL

d = m/v = 75.2 g / 89.2 mL = .84 g/mL

#13 m = 1450 g; V = 542 mL

d = m/v = 1450 g / 542 mL = 2.68 g/mL

#14 Scientific Method

1. Problem

2. Hypothesis

3. Experiment

4. Analyze Data

5. Conclusion/Retest

Gas Laws

Boyles Law P1V1 = P2V2 (inverse relationship)

When pressure increases volume decreasesWhen pressure decreases volume increases

Charles Law V1/T1 = V2/T2 (direct relationship)

When temperature increases volume increasesWhen temperature decreases volume decreases