1 st 6 weeks review test wednesday remember your pencil and calculator
TRANSCRIPT
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