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Properties of Matter II
Grade Nine Science Properties of Matter II
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Day 2: Handouts to Review
Matter concept map. Blackline master 1.2 “Properties of Matter”. Identify the property that best fits.
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1.5 Density Which is heavier: a 12 inch bar of gold or a pillow full of feathers? A 12 inch bar of gold weighs more than a pillow full of feathers even though the pillow takes up more space. Why? Gold is more dense. It has more matter per unit volume.
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1.5 Density Density is the amount of matter per unit volume of that matter. Density(D) = Mass(m) / Volume(V) If you know any two of the three variables (D,m, or V), you can solve for the third. m = D x V V = m / D
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Sample density question #1
If a rock has a mass of 49 g and occupies a volume of 7 cm3, what is the density? D = m / V = 49 g / 7 cm3 = 7.0 g/ cm3
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Solve these sample density questions
An object has a mass of 250 g and occupies a volume of 14.5 cm3, what is the density? A piece of wood occupies a volume of 46 cm3 and it has a mass of 100 g. What is the density of the wood? An unknown metal has a density of 2.6 g/cm3 and a mass of 15 g. How much volume does this piece of metal occupy? A sample of a particular liquid has a density of 6.85 kg/L and it occupies a volume of 3.4 L. How much does this particular sample weigh?
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Answer to question #1 D = m / V = 250 g / 14.5 cm3 = g/ cm3
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Answer to question #2 D = m / V = 100g / 46 cm3 = 2.17 g/ cm3
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Answer to question #3 D = m / V V = m / D = 15 g / 2.6 g/cm3
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Answer to question #4 D = m / V M = D x V = 6.85 kg/L x 3.4 L
** Note that volume can be expressed as cm3, m3, ml or even L.
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Approximate Densities of Some Common Materials
Substance Den sity kg/m3 g/cm3 Gold 19300 19.3 Silver 10500 10.5 Aluminum 2700 2.7 Ice 920 .92 Wood (birch) 660 .66 Distilled Water 1000 1.0
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Water is more dense than ice!
Why is this so important? It means that ice floats in water instead of sinking. Think of the poor fish if this was the not the case. Why is ice less dense than water? When water freezes, it expands which means that the same mass of water is spread over a greater volume.
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Homework Answer questions 2,3,4 on page 25 of the text.
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Day 3: Lab #1 Identifying Substances
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