REVIEW FOR UNIT ONE TEST- ANSWER KEY

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1 CHEMISTRY CP Name: Per. REVIEW FOR UNIT ONE TEST- ANSWER KEY Introductory Material: Scientific Method, Scientific Notation, Density 1. You will need to know the Metric Conversion Chart for this test. As a sample problem, give the numerical value for each of the prefixes: A. Mega 1,000,000 E. deci 0.1 B. Kilo 1,000 F. centi 0.01 C. Hecto 100 G. milli D. Deka Metric Conversions: Give the value of the following in the units indicated. a mm = m b. 8 Kg = 80,000 dg c. 35 dam = 0.35 Km d. 690 ml = L e..068 m = 68 mm f. 934 cm = hm g..55 L = 550 ml h Kg = 6,450,000 mg i. 75 mg = 0.075g 3. Be able to recognize which step of the scientific method is being described from different examples. Remember the basic steps are the following: a) observation, b) hypothesis, c) experimentation or testing, d) collecting or recording data, e) drawing conclusions. As an example, which step of the scientific method would each of the following be? (1) After years of testing and analyzing data a scientist determines that chemicals in a city s drinking water are causing an increase in the intelligence of children in the city. E- Drawing Conclusions (2) A biology student notices many frog tadpoles in a particular pond have two tails. A- Observation (3) The biology student prepares tables and graphs of all the information she has collected. D - Collecting or recording data (4) The biology student guesses that waste water from a sewage treatment plant which empties into the pond is causing the two tailed tadpoles. B - Hypothesis (5) The biology student obtains some freshly laid frog eggs and keeps some in the pond water and others in distilled water while waiting for them to hatch. C Experimentation or testing Problems: Calculate the following using correct units and significant digits. 4. A metal has a volume of 4.2 cm 3 and a mass of 16.0 g. What is its density? D = M/V = (16.0 g/ 4.2 cm 3 ) = 3.8 g/cm 3 5. An object has a mass of 75 g and a volume of 80.0 cm 3. What is its density? D = M/V = 75 g / 80.0 cm 3 ) =.94 g/cm 3 6. An object has a density of 4.5 g/cm 3 and a mass of 20.6 g. What is its volume? V = M/D = 20.6 g/ 4.5 g/cm 3 = 4.6 cm 3 7. Do the following multiplication and division problems and round off to the proper number of significant digits in the final answer. A = = 21 B = = 21 C. (23.002) (.0560) = =1.29 D. (84.00) (9001) = =756100

2 Perform the following operations and express your answer in proper scientific notation with the proper number of significant digits and proper units. 8. ( cm) ( cm) = cm 2 = cm 2 OR 1.5 cm g = g/dm 3 = g/dm dm ( Kg) ( m) = = kg m ( s) ( s) s m 3 = m m 12. Terms to Know: A. Conservation of Mass: Mass can neither be created nor destroyed. Their total amounts remain constant B. Chemistry (the four part definition): Chemistry is the study of matter, its structure, reactions with other matter and the energy involved C. Volume: Amount of space that an object occupies D. Density: Comparison of the amount of matter in an object to the space it occupies; mass per unit volume E. Weight: Pull of gravity on an object (can change depending on where you are) F. Mass: Amount of matter in an object (will not change depending on where you are) G. Accuracy: How closely individual measurements agree with the true value H. Precision: How close individual measurements agree with each other I. Mole: Number of basic particles of a substance in a gram molecular mass of that substance particles 13. Be able to read and interpret data off a line graph. VOLUME VS MASS READINGS FOR SAMPLES OF SUBSTANCE X a. What is the volume 20 grams of substance X would have? 10 ml Mass in grams b. What is the mass 15ml of substance X would have? 30 g c. Calculate the approximate slope of the line. y 2 -y 1 /x 2 -x 1 = 40-30/20-15 = Volume in ml d. Calculate the approximate density of the substance X. 2 g/ml Moles and Mole Conversions 1. What is Avogadro s Number? How many particles are in 1 Mole of carbon? particles 3. What is the mass that is equal to 1 Mole of Argon (Ar)? g 4. Complete the following Mass to Mole Conversions: DIVIDE BY MOLAR MASS a. 56 g of carbon (C) b..36 grams of helium (He) c. 225 grams of gold (Au) a. 56 g/12.01 g Mole = 4.7 M b..36 g/4.00 g Mole = M c. 225 g/197 g Mole = 1.14 M

3 5. Complete the following Mole to Mass Conversions: MULTIPLY BY MOLAR MASS a. 5 moles oxygen (O) b moles of neon (Ne) c..35 moles of potassium (K) g a. 5 Mole g g Mole Mole Mole 6. Complete the following Molecules to Mole Conversions: DIVIDE BY AVOGADRO S NUMBER a molecules of hydrogen b molecules of helium moleules moleules a Moles Moles b Moles Moles 7. Complete the following Mole to Molecules Conversions: MULTIPLY BY AVOGADRO S NUMBER a moles of carbon dioxide b moles of krypton a Moles ( molecules) b Moles ( molecules) 1 Mole 1 Mole molecules = molecules molecules = molecules Structure of Matter, Physical/Chemical Properties and States of Matter 1. Which of the following are chemical changes and which are physical changes? a. melting iron physical e. receiving a tattoo chemical b. burning hydrogen chemical f. tearing paper physical c. cutting a piece of wood physical g. exploding dynamite chemical d. digesting a Big Mac chemical h. evaporating salt water physical 2. Identify each of the following as being elements, compounds, homogeneous mixtures (solutions) or heterogeneous mixtures. ***Nonfat milk is homogeneous and Full fat milk is heterogeneous*** a. iron element h. ocean water homogeneous mixture b. dirt heterogeneous mixture i. lemonade homogeneous mixture c. rubbing alcohol compound j. water compound d. salt water homogeneous mixture k. corn syrup homogeneous mixture e. Italian salad dressing heterogeneous mixture l. granite heterogeneous mixture f. Nonfat milk homogeneous mixture g. mercury element 3. Identify each of the following as either chemical properties or physical properties. If they are physical properties identify them as either intensive or extensive. a. burning in air chemical f. volume physical extensive b. boiling point physical - intensive g. density physical - intensive c. malleability physical - intensive h. shape physical - intensive d. color physical - intensive i. odor physical - intensive e. melting point physical - intensive j. reactivity/stability chemical 4. Know the difference between the three states of matter in terms of shape, volumes, compressibility, and particle spacing. Solid Liquid Gas Description -Definite shape -Definite volume -No compressibility -Very little motion/ Vibrate in place -Takes shape of container -Constant volume -No compressibility -Particles slide/flow past each other -Fills container -Changeable volume -Easily compressed -Particles move freely 5. Know the four different methods for separating mixtures mentioned in the book: filtration, distillation, crystallization and chromatography. Filtration: a separation technique that uses a porous barrier to separate a solid from a liquid (example is sand and water mixture through filter paper and funnel) Crystallization: a separation technique that results from the formation of pure solid particles of a substance from a solution contained a dissolved substance (example is rock candy forming from sugar water)

4 Chromotography: a technique that separates components from a mixture, the mobile phase, as they travel across the surface of another material, stationary phase (example is black ink separates into different components as it spreads across filter paper) Distillation: a separation technique that is based on the differences in boiling points of the substances involved (example is distilling water to remove impurities) 6. Conservation of mass problems: a. From a process that splits water to form hydrogen and oxygen, a scientist collected 20.0 grams of hydrogen and 159 grams of oxygen. How much water was originally involved in the process? Water hydrogen + oxygen x grams = 20.0 grams grams x grams = 179 grams b. In a flask, 20.3 grams of aluminum reacted with grams of bromine to form aluminum bromide. After the reaction, no aluminum remained and 17 grams of bromine remained un-reacted. How many grams of bromine reacted? How many grams of aluminum bromide were formed? Aluminum + bromine aluminum bromide 20.3 grams grams = x grams 20.3 grams grams = grams Bromine (reacted) = Bromine (used) Bromine (remained) = g 17 g = 183 grams Aluminum bromide (formed) = Aluminum + Bromine (reacted) = 20.3 g g = grams 7. Percent by mass problems: a. A 156 gram sample of an unknown compound contains 25 grams of hydrogen. What is the percent by mass of hydrogen in the compound? mass of element 25 grams % by mass = 100 = 100 = 16% mass of compound 156 grams b. If 7.0 grams of substance X reacts with 21 grams of substance Y to form the compound XY, what is the percent by mass of each element in the compound? X + Y = XY 7.0 g + 21 g = 28 g % by mass X = % by mass Y = mass of element mass of compound mass of element mass of compound 7.0 grams 100 = 100 = 25% 28 grams 21 grams 100 = 100 = 75% OR (100% - 25% = 75% 28 grams c. Two unknown compounds are tested. Compound 1 contains 90.0 grams of hydrogen and 180 grams of oxygen. Compound 2 contains 45 grams of hydrogen and 90.0 grams of oxygen. Are the compounds the same? Compound 1: 90.0 grams grams = 270 grams 90.0 grams 100 = 33% hydrogen 100% - 33% = 67% oxygen 270 grams Compound 2: 45 grams grams = 135 grams 45 grams 100 = 33% hydrogen 100% - 33% = 67% oxygen 135 grams YES, THE COMPOUNDS ARE THE SAME 8. Law of Definite and Multiple Proportions a. A sample of a certain lead compound contains 6.46 g of lead for each gram of oxygen. A second sample has a mass of g and contains g of oxygen. Are the two samples the same? Compound 1: Set up a lead to oxygen ratio 6.46 grams = grams Compound 2: grams grams = grams lead Set up a lead to oxygen ratio grams =1.38 Because the ratios are NOT equal, these are NOT the same compound grams

5 b. Two different compounds are formed by the elements carbon and oxygen. The first compound contains 42.9% by mass carbon and 57.1% by mass oxygen. The second compound contains 27.3% by mass carbon and 72.7% by mass oxygen. Show that the data are consistent with the Law of Multiple Proportions. Compound 1: Set up a carbon to oxygen ratio 42.9 grams 57.1 grams = Compound 2: Set up a lead to oxygen ratio 27.3 grams 72.7 grams = Set up a ratio of Compound 1 to Compound 2: COMP 1 = = 2: 1 COMP

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