About Delphi Programming For Novice Developers

Hi! Im Zarko Gajic, your About.com Guide to Delphi Programming. Thats my picture at the top of the page (or maybe at the bottom). You can read my bio to learn more about who I am. I write feature articles and tutorials related to Delphi programming. I also gather links to other sites that have articles, tutorials, and important information on specific aspects of programming in the Delphi language. The purpose of this page is to orient newcomers with an overview of some or our special Delphi programming features. Embarcadero Technologies Delphi is an object-oriented, visual programming environment to develop 32 and 64 bit applications; with FireMonkey, Delphi is the fastest way to deliver ultra-rich and visually stunning native applications for Windows, Mac and iOS. If you are just entering the programming world, heres why you should consider learning Delphi: Why Delphi?. Also, dont miss Delphi History! If you are confused about different Delphi versions (Delphi Starter, Delphi XE2, RAD Studio), read the Flavors of Delphi article to easily pick your Delphi of choice. There is a lot of information on this site about Delphi programming; this site covers all aspects of Delphi development, including tutorials and articles, forum, language reference with examples, glossary, free code programs, custom components and much more. Let me help you find what youre looking for (and help your career by looking for the right Delphi job). Learn how Delphi can help you solve complex development problems to deliver high-performance, highly scalable applications ranging from Windows and database applications to mobile and distributed applications for the Internet. If you simply want to build a simple database application (accounting, CD/DVD album), for home use, Delphi will help you build it fast and with ease. Looking for something specific?You can search this Delphi Programming site or all of About.com for a specific programming task. Try it using the search box at the top of the page. Hint: Put phrases in double-quotation marks for better results (i.e. protected hack). If you are looking for more ways to find Delphi programming related materials, go see the Searching for Delphi article. True Beginners, Students, Newcomers ...For those who are new to Delphi, Ive prepared several free online courses designed to get you to a fast start. The free courses below are perfect for Delphi beginners as well as for those who want a broad overview of the art of programming with Delphi. Turbo Delphi Tutorial: For Novice and Non-ProgrammersA Beginners Guide to Delphi ProgrammingA Beginners Guide to Delphi Database ProgrammingA Beginners Guide to ASP.NET Web programming for Delphi developers Be sure not to miss the Delphi Tutorials and Online / Email Courses section. How to program in Delphi – what you need to know?This entire site is devoted to providing the tutorials and other resources needed to learn Delphi programming. There are several broad categories of Delphi programming tutorials to help you in your quest to learn how to create the best solutions fast. These include tutorials for the beginner as well as the more experienced developer, find them listed in A Beginners Guide to Delphi [enter Delphi topic]. If you are looking for free or/and shareware and commercial components, you’ll be happy to know I’ve prepared a dozen of Top Picks pages – where all the best third-party components, tools and Delphi books are collected and reviewed.

The Three Developmental Processes Are Biological - 1361 Words

Reading Assignment 1 The three developmental processes are Biological (Physical), Cognitive, and Socioemotional. The Biological developmental process focuses on the physical development of an individual, such as perceptual and motor capacities and changes in the body’s size, while the Cognitive process focuses on the cognitive development [memory, creativity, language, and knowledge]. The Socioemotional developmental process is focused on the changes in the individual’s psychosocial development, so it covers changes involving self-sufficiency and self-understanding, along with their morality and emotional communication. Each period of development includes pieces of the three developmental processes. Because of this, the developmental†¦show more content†¦- The Cognitive changes include the emergence of capacities such as perception and intellect. - The Socioemotional changes that take place include the child making its first connection to another human being. †¢ Early Childhood: This period is from two to six years. - The Biological changes that occur during this period of time include the individual’s body growing and becoming taller and thinner. - The Cognitive changes occurring include the child becoming more self-sufficient and the capacities of thought and language begin to expand. - The Socioemotional changes taking place include the child growing in ways of morality, self-understanding, and in their relationships with peers. †¢ Middle Childhood: This period takes place from 6 to 11 years. - The Biological changes taking place during this period include advancement in athletic abilities. - The Cognitive changes that occur in this period include the child becoming more logical in their though processes and their basic literacy skills come more naturally to them. - The Socioemotional changes that take place include the child’s self-understanding and morality becoming more evident, as well as more advanced. It also shows progress in their ability to create and maintain friendships. †¢ Adolescence: This period spans from 11 to 18 years. - The Biological changes taking place include the occurrence of Puberty which leads to changes in the body, such as, the body growing larger and becoming â€Å"adult-sized†Show MoreRelatedCognitive Psychology And The Management And Treatment Of Mental Illness1352 Words   |  6 Pagesdifferences. AC1.1, AC3.1 Biological psychology looks at the biological aspects of behaviour. It looks at how the brain s structure, chemistry, activity and genetic make-up etc. relates to behaviour. Cognitive psychology focuses on the way the brain processes information, how people perceive, understand, make decisions about and remember information. Cognitive psychologists would put information in to be processed and then see what the brain does with it. To compare the biological psychologist wouldRead MoreThe Problem Of Human Development1536 Words   |  7 Pagesbehind it. Experts have not been able to discover any exact influences of normal human development, nor have they been able to uncover the exact causes of developmental abnormalities (Disabilities). What they have found can seem quite confusing and has caused much debate among those trying to tackle the answer to the question. Is ones developmental trajectory determined by the highness of their genetics or what they experience from the environm ent to which they are raised? Because of the spectrumRead MoreIntroduction The three major motor developmental theories are maturation, sensory processing and1400 Words   |  6 PagesIntroduction The three major motor developmental theories are maturation, sensory processing and dynamic. Maturation refers to the biological growth processes that begin at conception and lead to the maturity of the body. These innate (inborn) changes occur in the body, brain and nervous system automatically in an orderly sequence at about the same time for most children. Because these changes are innate, they are not influenced by environmental experiences. Many of our basic abilities are closelyRead MoreSchizophrenia/Psychosis/Life Span948 Words   |  4 Pagestranslates as split mind and the psychological changes can be so profound that the affected individual is thrust into a world that bears little resemblance to everyday experience. The person with schizophrenia lives in an internal world marked by thought processes that have gone awry; delusions, hallucinations, and generally disordered thinking become the norm. Hansell and Damour (2005) states: Psychosis is a state of being profoundly out of touch with reality. Psychotic individuals may experience hallucinationsRead MoreComparing Psychological Theories Essay856 Words   |  4 Pages(Gottlieb 2002). The Biological theory emphasizes the influence of biology on our behavior. Psychologists assume that our mental processes, that is our thoughts, fantasies, and dreams, are made possible by the nervous system. They point especially to its key component, the brain (Lickliter Honeycutt 2003). Biologically oriented psychologists look for the connections between events in the brain, such as activity of the brain cells, and behavior and mental processes. All of these theoriesRead More The Developing Adolescent Essay1402 Words   |  6 PagesDevelopmental theories are a group of ideas, assumptions, and generalizations that interpret and illuminate the thousands of observations that have been made about human growth. In this way, developmental theories provide a framework for explaining the patterns and problems of development (Berger, 2008 p33). Adolescence represents one of the most critical developmental periods in life. It is a time of profound changes on all levels. The importance of both family and peers during these years is alsoRead More The Nature of Child Development Essay1172 Words   |  5 Pages the former approach has found its roots in the biological structure of the human organism which considers our development as a series of stages. However, referring to human development only as continuous or gradual would produce inconsistent and insufficient understanding of the developmental process .As a result, combining idea s of both viewpoints(continuity and phases) will provide a more explicit of understanding of the developmental processes and their final aim which is to create individual`sRead MoreTheories And Theories Of Development996 Words   |  4 Pagesformulate hypothesis, or testable answers, to why questions about behavior. At the broadcast level there are three families of theories including psychoanalytic theories, learning theories, and cognitive theories. These theories attempt to provide developmentalists with compressive explanations for just about every fact of human development. Additionally, theories that deal with the biological foundations of development and interactions between these factors and development and interactions betweenRead MoreThe Impact Of Sensory Processing Disorder On Development And Development1500 Words   |  6 PagesIntroduction The human life span is from conception to death and can be divided into eight different developmental periods. This essay will focus on the early childhood period that ranges from two to five or six years of age. Firstly, this essay will describe the typical developmental milestones during early childhood and how they play an important role in each developmental stage that follows. Secondly, it will analyse how the home and educational environments influence early childhood. FinallyRead MoreUnderstanding Piaget Theory And Information Processing Theory1208 Words   |  5 PagesThe study of Cognitive theories has many different aspects that have been debated many years ago. Developmental psychologists try to explain cognitive development approaches which describe the process of human s thought. One of the developmental psychologist who studied on the area of cognitive was Jean Piaget. Jean Piaget a Swiss psychologist was the first developmental researcher who has extensive research on cognitive development. In addition, the revolution of Jean Piaget’s cognitive theory

General Chemistry free essay sample

For a person who weighs 100 pounds or more, three significant figures are typically used to report the weight (given to the whole pound), although people often round to the nearest unit of 10, which may result in reporting the weight with two significant figures (for example, 170 pounds). 165 pounds rounded to two significant figures would be reported as 1. 7 x 102 pounds. b. Copyright  © Houghton Mifflin Company. All rights reserved. 4 c. Chapter 1: Chemistry and Measurement For example, 165 lb weighed on a scale that can measure in 100-lb increments would be 200 lb. Using the conversion factor 1 lb = 0. 536 kg, 165 lb is equivalent to 74. 8 kg. Thus, on a scale that can measure in 50-kg increments, 165 lb would be 50 kg. 1. 3. a. b. c. d. If your leg is approximately 32 inches long, this would be equivalent to 0. 81 m, 8. 1 dm, or 81 cm. One story is approximately 10 feet, so three stories is 30 feet. This would be equivalent to approximately 9 m. Normal body temperature is 98. 6 °F, or 37. 0 °C. Thus, if your body temperature were 39 °C (102 °F), you would feel as if you had a moderate fever. Room temperature is approximately 72 °F, or 22 °C. Thus, if you were sitting in a room at 23 °C (73 °F), you would be comfortable in a short-sleeve shirt. . 4. Gold is a very unreactive substance, so comparing physical properties is probably your best option. However, color is a physical property you cannot rely on in this case to get your answer. One experiment you could perform is to determine the densities of the metal and the chunk of gold. You could measure the mass of the nugget on a balance and the volume of the nugget by water displacement. Using this information, you could calculate the density of the nugget. Repeat the experiment and calculations for the sample of gold. If the nugget is gold, the two densities should be equal and be 19. g/cm3. Also, you could determine the melting points of the metal and the chunk of pure gold. The two melting points should be the same (1338 K) if the metal is gold.  ¦ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS 1. 1. One area of technology that chemistry has changed is the characteristics of materials. The liquidcrystal displays (LCDs) in devices such as watches, cell phones, computer monitors, and televisions are materials made of molecules designed by chemists. Electronics and communications have been transformed by the development of optical fibers to replace copper wires. In biology, chemistry has changed the way scientists view life. Biochemists have found that all forms of life share many of the same molecules and molecular processes. 1. 2. An experiment is an observation of natural phenomena carried out in a controlled manner so that the results can be duplicated and rational conclusions obtained. A theory is a tested explanation of basic natural phenomena. They are related in that a theory is based on the results of many experiments and is fruitful in suggesting other, new experiments. Also, an experiment can disprove a theory but can never prove it absolutely. A hypothesis is a tentative explanation of some regularity of nature. Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 5 1. 3. Rosenberg conducted controlled experiments and noted a basic relationship that could be stated as a hypothesis—that is, that certain platinum compounds inhibit cell division. This led him to do new experiments on the anticancer activity of these compounds. 1. 4. Matter is the general term for the material things around us. It is whatever occupies space and can be perceived by our senses. Mass is the quantity of matter in a material. The difference between mass and weight is that mass remains the same wherever it is measured, but weight is proportional to the mass of the object divided by the square of the distance between the center of mass of the object and that of the earth. 1. 5. The law of conservation of mass states that the total mass remains constant during a chemical change (chemical reaction). To demonstrate this law, place a sample of wood in a sealed vessel with air, and weigh it. Heat the vessel to burn the wood, and weigh the vessel after the experiment. The weight before the experiment and that after it should be the same. 1. 6. Mercury metal, which is a liquid, reacts with oxygen gas to form solid mercury(II) oxide. The color changes from that of metallic mercury (silvery) to a color that varies from red to yellow depending on the particle size of the oxide. 1. 7. Gases are easily compressible and fluid. Liquids are relatively incompressible and fluid. Solids are relatively incompressible and rigid. 1. 8. An example of a substance is the element sodium. Among its physical properties: It is a solid, and it melts at 98 °C. Among its chemical properties: It reacts vigorously with water, and it burns in chlorine gas to form sodium chloride. 1. . An example of an element: sodium; of a compound: sodium chloride, or table salt; of a heterogeneous mixture: salt and sugar; of a homogeneous mixture: sodium chloride dissolved in water to form a solution. 1. 10. A glass of bubbling carbonated beverage with ice cubes contains three phases: gas, liquid, and solid. 1. 11. A compound may be decomposed by chemical reactio ns into elements. An element cannot be decomposed by any chemical reaction. Thus, a compound cannot also be an element in any case. 1. 12. The precision refers to the closeness of the set of values obtained from identical measurements of a quantity. The number of digits reported for the value of a measured or calculated quantity (significant figures) indicates the precision of the value. Copyright  © Houghton Mifflin Company. All rights reserved. 6 Chapter 1: Chemistry and Measurement 1. 13. Multiplication and division rule: In performing the calculation 100. 0 x 0. 0634 ? 25. 31, the calculator display shows 0. 2504938. We would report the answer as 0. 250 because the factor 0. 0634 has the least number of significant figures (three). Addition and subtraction rule: In performing the calculation 184. 2 + 2. 324, the calculator display shows 186. 24. Because the quantity 184. 2 has the least number of decimal places (one), the answer is reported as 186. 5. 1. 14. An exact number is a number that arises when you count items or sometimes when you define a unit. For example, a foot is defined to be 12 inches. A measured number is the result of a comparison of a physical quantity with a fixed standard of measurement. For example, a steel rod measures 9. 12 centimeters, or 9. 12 times the standard centimeter unit of measurement. 1. 15. For a given unit, the SI system uses prefixes to obtain units of different sizes. Units for all other possible quantities are obtained by deriving them from any of the seven base units. You do this by using the base units in equations that define other physical quantities. 1. 16. An absolute temperature scale is a scale in which the lowest temperature that can be attained theoretically is zero. Degrees Celsius and kelvins have units of equal and are related by the formula tC = (TK ? 273. 15 K) x 1 °C 1K 1. 17. The density of an object is its mass per unit volume. Because the density is characteristic of a substance, it can be helpful in identifying it. Density can also be useful in determining whether a substance is pure. It also provides a useful relationship between mass and volume. 1. 18. Units should be carried along because (1) the units for the answers will come out in the calculations, and (2), if you make an error in arranging factors in the calculation, this will become apparent because the final units will be nonsense. 1. 19. The answer is c, three significant figures. 1. 20. The answer is a, 4. 43 x 102 mm. 1. 21. The answer is e, 75 mL. 1. 22. The answer is c, 0. 23 mg. Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement  ¦ 1. 23. ANSWERS TO CONCEPT EXPLORATIONS a. First, check the physical appearance of each sample. Check the particles that make up each sample for consistency and hardness. Also, note any odor. Then perform on each sample some experiments to measure physical properties such as melting point, density, and solubility in water. Compare all of these results a nd see if they match. It is easier to prove that the compounds were different by finding one physical property that is different, say different melting points. To prove the two compounds were the same would require showing that every physical property was the same. Of the properties listed in part a, the melting point would be most convincing. It is not difficult to measure, and it is relatively accurate. The density of a powder is not as easy to determine as the melting point, and solubility is not reliable enough on its own. No. Since neither solution reached a saturation point, there is not enough information to tell if there was a difference in behavior. Many white powders dissolve in water. Their chemical compositions are not the same. b. c. d. 1. 24. Part 1 a. b. c. d. e. 3 g + 1. 4 g + 3. 3 g = 7. 7 g = 8 g First, 3 g + 1. 4 g = 4. g = 4 g. Then, 4 g + 3. 3 g = 7. 3 g = 7 g. Yes, the answer in part a is more accurate. When you round off intermediate steps, you accumulate small errors and your answer is not as accurate. The answer 29 g is correct. This answer is incorrect. It should be 3 x 101 with only one significant figure in the answer. The student probably applied the rule for addition (instead of for multiplication) after the first step. The answer 28. 5 g is correct. Don’t round off intermediate answers. Indicate the round-off position after each step by underlining the least significant digit. The calculated answer is incorrect. It should be 11 cm3. The answer given has too many significant figures. There is also a small round off error due to using a rounded-off value for the density. This is a better answer. It is reported with the correct number of significant figures (three). It can be improved by using all of the digits given for the density. V = f. g. Part 2 a. b. c. d. 10 ball bearings 1. 234 g 1 cm3 x x = 3. 90889 = 3. 909 cm3 1 3. 1569 g 1 ball bearing There was no rounding off of intermediate steps; all the factors are as accurate as possible. Copyright  © Houghton Mifflin Company. All rights reserved. 8 Chapter 1: Chemistry and Measurement  ¦ 1. 25. ANSWERS TO CONCEPTUAL PROBLEMS a. b. Two phases: liquid and solid. Three phases: liquid water, solid quartz, and solid seashells. 1. 26. If the material is a pure compound, all samples should have the same melting point, the same color, and the same elemental composition. If it is a mixture, these properties should differ depending on the composition. 1. 27. a. You need to establish two points on the thermometer with known (defined) temperatures— for example, the freezing point (0 °C) and boiling point (100 °C) of water. You could first immerse the thermometer in an ice-water bath and mark the level at this point as 0 °C. Then, immerse the thermometer in boiling water, and mark the level at this point as 100 °C. As long as the two points are far enough apart to obtain readings of the desired accuracy, the thermometer can be used in experiments. You could make 19 evenly spaced marks on the thermometer between the two original points, each representing a difference of 5 °C. You may divide the space between the two original points into fewer spaces as long as you can read the thermometer to obtain the desired accuracy. a. b. c. b. 1. 8. 1. 29. a. To answer this question, you need to develop an equation that converts between  °F and  °YS. To do so, you need to recognize that one degree on the Your Scale does not correspond to one degree on the Fahrenheit scale and that ? 100 °F corresponds to 0 ° on Your Scale (different â€Å"zero† points). As stated in the problem, in the desired r ange of 100 Your Scale degrees, there are 120 Fahrenheit degrees. Therefore, the relationship can be expressed as 120 °F = 100 °YS, since it covers the same temperature range. Now you need to â€Å"scale† the two systems so that they correctly convert from one scale to the other. You could set up an equation with the known data points and then employ the information from the relationship above. Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 9 For example, to construct the conversion between  °YS and  °F, you could perform the following steps: Step 1:  °F =  °YS Not a true statement, but one you would like to make true. Step 2:  °F =  °YS x 120 °F 100 °YS This equation takes into account the difference in the size between the temperature unit on the two scales but will not give you the correct answer because it doesn’t take into account the different zero points. Step 3: By subtracting 100 °F from your equation from Step 2, you now have the complete equation that converts between  °F and  °YS.  °F = ( °YS x 120 °F ) ? 100 °F 100 °YS b. Using the relationship from part a, 66 °YS is equivalent to (66 °YS x 120 °F ) ? 100 °F = ? 20. 8 °F = ? 21 °F 100 °YS 1. 30. Some physical properties you could measure are density, hardness, color, and conductivity. Chemical properties of sodium would include reaction with air, reaction with water, reaction with chlorine, reaction with acids, bases, etc. 1. 31. The empty boxes are identical, so they do not contribute to any mass or density difference. Since the edge of the cube and the diameter of the sphere are identical, they will occupy the same volume in each of the boxes; therefore, each box will contain the same number of cubes or spheres. If you view the spheres as cubes that have been rounded by removing wood, you can conclude that the box containing the cubes must have a greater mass of wood; hence, it must have a greater density. 1. 32. a. b. Since the bead is less dense than any of the liquids in the container, the bead will float on top of all the liquids. First, determine the density of the plastic bead. Since density is mass divided by volume, you get d = m 3. 2 x 10-2 g = = 0. 911 g/mL = 0. 91 g/mL V 0. 043 mL Thus, the glass bead will pass through the top three layers and float on the ethylene glycol layer, which is more dense. c. Since the bead sinks all the way to the bottom, it must be more dense than 1. 114 g/mL. Copyright  © Houghton Mifflin Company. All rights reserved. 10 1. 33. a. b. Chapter 1: Chemistry and Measurement A paper clip has a mass of about 1 g. Answers will vary depending on your particular sample. Keeping in mind that the SI unit for mass is kg, the approximate weights for the items presented in the problem are as follows: a grain of sand, 1 x 10? kg; a paper clip, 1 x 10? 3 kg; a nickel, 5 x 10? 3 kg; a 5. 0-gallon bucket of water, 2. 0 x 101 kg; a brick, 3 kg; a car, 1 x 103 kg. 1. 34. When taking measurements, never throw away meaningful information even if there is some uncertainty in the final digit. In this case, you are certain that the nail is between 5 and 6 cm. The uncertain, yet still important, digit is between the 5 and 6 cm measurements. You can estimate with reasonable precision that it is about 0. 7 cm from the 5 cm mark, so an acceptable answer would be 5. 7 cm. Another person might argue that the length of the nail is closer to 5. cm, which is also acceptable given the precision of the ruler. In any case, an answer of 5. 7 or 5. 8 should provide usefu l information about the length of the nail. If you were to report the length of the nail as 6 cm, you would be discarding potentially useful length information provided by the measuring instrument. If a higher degree of measurement precision were needed (more significant figures), you would need to switch to a more precise ruler—for example, one that had mm markings. 1. 35. a. The number of significant figures in this answer follows the rules for multiplication and division. Here, the measurement with the fewest significant figures is the reported volume 0. 310 m3, which has three. Therefore, the answer will have three significant figures. Since Volume = L x W x H, you can rearrange and solve for one of the measurements, say the length. L = V 0. 310 m 3 = = 0. 83496 m = 0. 835 m W x H (0. 7120 m) (0. 52145 m) b. The number of significant figures in this answer follows the rules for addition and subtraction. The measurement with the least number of decimal places is the result 1. 509 m, which has three. Therefore, the answer will have three decimal places. Since the result is the sum of the three measurements, the third length is obtained by subtracting the other two measurements from the total. Length = 1. 509 m ? 0. 7120 m ? 0. 52145 m = 0. 27555 m = 0. 276 m 1. 36. The mass of something (how heavy it is) depends on how much of the item, material, substance, or collection of things you have. The density of something is the mass of a specific amount (volume) of an item, material, substance, or collection of things. You could use 1 kg of feathers and 1 kg of water to illustrate that they have the same mass yet have very different volumes; therefore, they have different densities. Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 11  ¦ SOLUTIONS TO PRACTICE PROBLEMS Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multistep problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. . 37. By the law of conservation of mass: Mass of sodium carbonate + mass of acetic acid solution = mass of contents of reaction vessel + mass of carbon dioxide Plugging in gives 15. 9 g + 20. 0 g = 29. 3 g + mass of carbon dioxide Mass of carbon dioxide = 15. 9 g + 20. 0 g ? 29. 3 g = 6. 6 g 1. 38. By the law of conservation of mass: Mass of iron + mass of acid = mass of contents of beaker + mass of hydrogen Plugging in gives 5. 6 g + 15. 0 = 20. 4 g + mass of hydrogen Mass of hydrogen = 5. 6 g + 15. 0 g ? 20. 4 g = 0. 2 g 1. 39. By the law of conservation of mass: Mass of zinc + mass of sulfur = mass of zinc sulfide Rearranging and plugging in give Mass of zinc sulfide = 65. 4 g + 32. 1 g = 97. 5 g For the second part, let x = mass of zinc sulfide that could be produced. By the law of conservation of mass: 20. 0 g + mass of sulfur = x Write a proportion that relates the mass of zinc reacted to the mass of zinc sulfide formed, which should be the same for both cases. mass zinc 65. 4 g 20. 0 g = = mass zinc sulfide 97. 5 g x Solving gives x = 29. 81 g = 29. 8 g Copyright  © Houghton Mifflin Company. All rights reserved. 12 Chapter 1: Chemistry and Measurement 1. 0. By the law of conservation of mass: Mass of aluminum + mass of bromine = mass of aluminum bromide Plugging in and solving give 27. 0 g + Mass of bromine = 266. 7 g Mass of bromine = 266. 7 g ? 27. 0 g = 239. 7 g For the second part, let x = mass of bromine that reacts. By the law of conservation of mass: 15. 0 g + x = mass of aluminum bromide Writ e a proportion that relates the mass of aluminum reacted to the mass of bromine reacted, which should be the same for both cases. mass aluminum 27. 0 g 15. 0 g = = x mass bromine 239. 7 g Solving gives x = 133. 1 g = 133 g 1. 41. a. b. c. d. 1. 42. a. b. c. d. 1. 43. a. b. c. d. 1. 44. a. . c. Physical change Chemical change Chemical change Physical change Physical change Chemical change Physical change Solid Solid Solid Liquid Solid Liquid Gas Solid Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 13 d. Physical change 1. 45. Physical change: Liquid mercury is cooled to solid mercury. Chemical changes: (1) Solid mercury oxide forms liquid mercury metal and gaseous oxygen; (2) glowing wood and oxygen form burning wood (form ash and gaseous products). 1. 46. Physical changes: (1) Solid iodine is heated to gaseous iodine; (2) gaseous iodine is cooled to form solid iodine. Chemical change: Solid iodine and zinc metal are ignited to form a white powder. 1. 47. a. b. c. d. e. 1. 48. a. b. c. d. e. Physical property Chemical property Physical property Chemical property Physical property Physical property Chemical property Physical property Physical property Chemical property 1. 49. Physical properties: (1) Iodine is solid; (2) the solid has lustrous blue-black crystals; (3) the crystals vaporize readily to a violet-colored gas. Chemical properties: (1) Iodine combines with many metals, such as with aluminum to give aluminum iodide. 1. 50. Physical properties: (1) is a solid; (2) has an orange-red color; (3) has a density of 11. 1 g/cm3; (4) is insoluble in water. Chemical property: Mercury(II) oxide decomposes when heated to give mercury and oxygen. 1. 51. a. b. c. Physical process Chemical reaction Physical process Copyright  © Houghton Mifflin Company. All rights reserved. 14 d. e. 1. 52. a. b. c. d. e. 1. 53. a. b. c. d. 1. 54. a. b. c. d. 1. 55. a. b. c. d. 1. 56. a. b. c. d. Chapter 1: Chemistry and Measurement Chemical reaction Physical process Chemical reaction Physical process Physical process Physical process Chemical reaction Solution Substance Substance Homogeneous or Heterogeneous mixture Heterogeneous mixture Substance Solution Substance A pure substance with two phases present, liquid and gas. A mixture with two phases present, solid and liquid. A pure substance with two phases present, solid and liquid. A mixture with two phases present, solid and solid. A mixture with two phases present, solid and liquid. A mixture with two phases present, solid and liquid. A mixture with two phases present, solid and solid. A pure substance with two phases present, liquid and gas. Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 15 1. 57. a. b. c. d. e. f. 1. 58. a. b. c. d. e. f. three four six four four four six three four five three four 1. 59. 40,000 km = 4. 0 x 104 km 1. 60. 150,000,000 km = 1. 50 x 108 km 1. 61. a. b. c. d. 1. 62. a. b. c. d. 0. 871 x 0. 57 = 0. 08456 = 0. 085 5. 871 8. 937 ? 8. 930 = 0. 007 8. 937 + 8. 930 = 17. 867 0. 00015 x 54. 6 + 1. 002 = 0. 00819 + 1. 002 = 1. 0101 = 1. 010 8. 71 x 0. 0301 = 8. 457 = 8. 5 0. 031 0. 71 + 92. 2 = 92. 91 = 92. 9 934 x 0. 00435 + 107 = 4. 0629 + 107 = 111. 06 = 111 (847. 89 ? 847. 73) x 14673 = 0. 16 x 14673 = 2347 = 2. 3 x 103 Copyright  © Houghton Mifflin Company. All rights reserved. 16 Chapter 1: Chemistry and Measurement 1. 63. The volume of the first sphere is V1 = (4/3)? r3 = (4/3)? x (5. 10 cm)3 = 555. 64 cm3 The volume of the second sphere is V2 = (4/3)? r3 = (4/3)? x (5. 00 cm)3 = 523. 60 cm3 The difference in volume is V1 ? V2 = 555. 64 cm3 ? 523. 60 cm3 = 32. 04 cm3 = 32 cm3 1. 64. The length of the cylinder between the two marks is l = 3. 50 cm ? 3. 10 cm = 0. 40 cm The volume of iron contained between the marks is V = ? r2l = ? x (1. 500 cm)2 x 0. 40 cm = 2. 82 cm3 = 2. 8 cm3 1. 65. a. b. c. d. 1. 66. a. b. c. d. 1. 67. a. b. c. d. 1. 68. a. b. c. d. 6. 20 km = 6. 0 x 103 m 1. 98 ns = 1. 98 x 10? 9 s 2. 54 cm = 2. 54 x 10? 2 m 5. 23  µg = 5. 23 x 10? 6 g 6. 15 ps = 6. 15 x 10? 12 s 3. 781  µm = 3. 781 x 10? 6 m 1. 546 A = 1. 546 x 10? 10 m 9. 7 mg = 9. 7 x 10? 3 g 4. 851 x 10? 6 g = 4. 851  µg 3. 16 x 10? 2 m = 3. 16 cm 2. 591 x 10? 9 s = 2. 591 ns 8. 93 x 10? 12 g = 8. 93 pg 5. 89 x 10? 12 s = 5. 89 ps 0. 2010 m = 20. 10 cm 2. 560 x 10? 9 g = 2. 560 ng 6. 05 x 103 m = 6. 05 km Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 17 1. 69. a. b. c. d. tC = tC = 5 °C 5 °C x (tF ? 32 °F) = x (68 °F ? 32 °F) = 20. 0 °C = 20.  °C 9 °F 9 °F 5 °C 5 °C x (tF ? 2 °F) = x (? 23 °F ? 32 °F) = ? 30. 55 °C = ? 31 °C 9 °F 9 °F 9 °F 9 °F ) + 32 °F = (26 °C x ) + 32 °F = 78. 8 °F = 79 °F 5 °C 5 °C 9 °F 9 °F ) + 32 °F = (? 70 °C x ) + 32 °F = ? 94. 0 °F = ? 94 °F 5 °C 5 °C tF = (tC x tF = (tC x 1. 70. a. b. c. d. tC = tC = 5 °C 5 °C x (tF ? 32 °F) = x (51 °F ? 32 °F) = 10. 555 °C = 11 °C 9 °F 9 °F 5 °C 5 °C x (tF ? 32 °F) = x (? 7 °F ? 32 °F) = ? 21. 6 °C = ? 22 °C 9 °F 9 °F 9 °F 9 °F ) + 32 °F = (? 41 °C x ) + 32 °F = ? 41. 8 °F = ? 42 °F 5 °C 5 °C tF = (tC x tF = (tC x 9 °F 9 °F ) + 32 °F = (22 °C x ) + 32 °F = 71. 6 °F = 72 °F 5 °C 5 °C 1. 71. tF = (tC x 9 °F 9 °F ) + 32 °F = (? 21. 1 °C x ) + 32 °F = ? 5. 98 °F = ? 6. 0 °F 5 °C 5 °C 1. 72. F = (tC x 9 °F 9 °F ) + 32 °F = (? 196 °C x ) + 32 °F = ? 320. 8 °F = ? 321 °F 5 °C 5 °C 1. 73. d = m 12. 4 g = = 7. 560 g/cm3 = 7. 56 g/cm3 V 1. 64 cm3 m 17. 84 g = = 0. 7136 g/mL = 0. 714 g/mL V 25. 0 mL 1. 74. d = 1. 75. First, determine the density of the liquid. d = m 6. 71 g = = 0. 7894 = 0. 79 g/mL V 8. 5 mL The density is closest to ethanol (0. 789 g/cm3). Copyright  © Houghton Mifflin Company. All rights reserved. 18 Chapter 1: Chemistry and Measurement 1. 76. First, determine the density of the mineral sample. d = m 59. 5 g = = 7. 531 = 7. 5 g/cm3 3 V 7. 9 cm The density is closest to cassiterite (6. 99 g/cm3). 1. 77. The mass of platinum is obtained as follows. Mass = d x V = 21. 4 g/cm3 x 5. 9 cm3 = 126 g = 1. 3 x 102 g 1. 78. The mass of gasoline is obtained as follows. Mass = d x V = 0. 70 g/mL x 43. 8 mL = 30. 66 g = 31 g 1. 79. The volume of ethanol is obtained as follows. Recall that 1 mL = 1 cm3. Volume = m 19. 8 g = = 25. 09 cm3 = 25. 1 cm3 = 25. 1 mL d 0. 789 g/cm3 1. 80. The volume of bromine is obtained as follows. Volume = m 88. 5 g = 28. 54 mL = 28. 5 mL = d 3. 10 g/mL 1. 81. Since 1 kg = 103 g, and 1 mg = 10? 3 g, you can write 0. 480 kg x 103 g 1 mg = 4. 80 x 105 mg x 1 kg 10-3 g 1. 82. Since 1 mg = 10? 3 g, and 1  µg = 10? g, you can write 501 mg x 10-3 g 1 ? g = 5. 01 x 105  µg x 1 mg 10-6 g 1. 83. Since 1 nm = 10? 9 m, and 1 cm = 10? 2 m, you can write 555 nm x 1 cm 10-9 m = 5. 55 x 10? 5 cm x -2 10 m 1 nm 1. 84. Since 1 A = 10? 10 m, you can write 0. 96 A x 10-10 m = 9. 6 x 10? 11 m 1A Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 19 1. 85. Since 1 km = 103 m, you can write ? 103 m ? 3. 73 x 10 km x ? ? ? 1 km ? 8 3 3 = 3. 73 x 1017 m3 Now, 1 dm = 10? 1 m. Also, note that 1 dm3 = 1 L. Therefore, you can write ? 1 dm ? 3. 73 x 1017 m3 x ? -1 ? = 3. 73 x 1020 dm3 = 3. 73 x 1020 L ? 10 m ? 3 1. 6. 1  µm = 10? 6 m, and 1 dm = 10? 1 m. Also, note that 1 dm3 = 1 L. Therefore, you can write ? 10-6 m ? 1. 3  µm x ? ? ? 1 ? m ? 3 3 ? 1 dm ? x ? -1 ? = 1. 3 x 10? 15 dm3 = 1. 3 x 10? 15 L ? 10 m ? 3 1. 87. 3. 58 short ton x 2000 lb 16 oz 1g x x = 3. 248 x 106 g = 3. 25 x 106 g 1 short ton 1 lb 0. 03527 oz 1. 88. 3. 15 Btu x 252. 0 cal 4. 184 J x = 3321 J = 3. 32 x 103 J 1 Btu 1 cal 6 ft 12 in. 2. 54 x 10-2 m = 4434. 8 m = 4. 435 x 103 m x x 1 fathom 1 ft 1 in. 42 gal 4 qt 9. 46 x 10-4 m3 = 2. 066 x 109 m3 = 2. 1 x 109 m3 x x 1 qt 1 barrel 1 gal 3 1. 89. 2425 fathoms x 1. 90. 1. 3 x 1010 barrels x 1L ? 2. 54 cm ? 1. 91. (20. in. )(20. 0 in. )(10. 0 in. ) x ? = 65. 54 L = 65. 5 L ? x 1000 cm3 ? 1 in. ? ? 1000 m ? 25 worms 1. 92. (1. 00 km)(2. 0 km)(1 m) x ? = 5. 00 x 107 = 5. 0 x 107 worms ? x 3 1m ? 1 km ? 2 Copyright  © Houghton Mifflin Company. All rights reserved. 20 Chapter 1: Chemistry and Measurement  ¦ SOLUTIONS TO GENERAL PROBLEMS 1. 93. From the law of conservation of mass, Mass of sodium + mass of water = mass of hydrogen + mass of solution Substituting, you obtain 19. 70 g + 126. 22 g = mass of hydrogen + 145. 06 g or, Mass of hydrogen = 19. 70 g + 126. 22 g ? 145. 06 g = 0. 86 g Thus, the mass of hydrogen produced was 0. 86 g. 1. 94. From the law of conservation of mass, Mass of tablet + mass of acid solution = mass of carbon dioxide + mass of solution Substituting, you obtain 0. 853 g + 56. 519 g = mass of carbon dioxide + 57. 152 g Mass of carbon dioxide = 0. 853 g + 56. 519 g ? 57. 152 g = 0. 220 g Thus, the mass of carbon dioxide produced was 0. 220 g. 1. 95. From the law of conservation of mass, Mass of aluminum + mass of iron(III) oxide = mass of iron + mass of aluminum oxide + mass of unreacted iron(III) oxide 5. 40 g + 18. 50 g = 11. 17 g + 10. 20 g + mass of iron(III) oxide unreacted Mass of iron(III) oxide unreacted = 5. 40 g + 18. 50 g ? 11. 17 g ? 0. 20 g = 2. 53 g Thus, the mass of unreacted iron(III) oxide is 2. 53 g. 1. 96. From the law of conservation of mass, Mass of sodium bromide + mass of chlorine reacted = mass of bromine + mass of sodium chloride 20. 6 g + mass of chlorine reacted = 16. 0 g + 11. 7 g Mass of chlorine reacted = 16. 0 g + 11. 7 g ? 20. 6 g = 7. 1 g Thus, the mass of chlorine t hat reacted is 7. 1 g. 1. 97. 53. 10 g + 5. 348 g + 56. 1 g = 114. 54 g = 114. 5 g total 1. 98. 68. 1 g + 58. 2 g + 5. 279 g = 131. 579 g = 131. 6 g total Copyright  © Houghton Mifflin Company. All rights reserved. Chapter 1: Chemistry and Measurement 21 1. 99. a. b. c. d. 1. 00. a. b. c. d. Physical Chemical Physical Chemical Chemical Physical Physical Chemical 1. 101. Compounds always contain the same proportions of the elements by mass. Thus, if we let X be the proportion of iron in a sample, we can calculate the proportion of iron in each sample as follows. Sample A: Sample B: Sample C: X = X = X = mass of iron 1. 094 g = = 0. 72068 = 0. 7207 mass of sample 1. 518 g mass of iron 1. 449 g = = 0. 70476 = 0. 7048 mass of sample 2. 056 g mass of iron 1. 335 g = = 0. 71276 = 0. 7128 mass of sample 1. 873 g Since each sample has a different proportion of iron by mass, the material is not a compound. . 102. Compounds always contain the same proportions of the elements by mass. Thus, if we let X be the proportion of mercury in a sample, we can calculate the proportion of mercury in each sample as follows. Sample A: Sample B: Sample C: X = X = X = mass of mercury 0. 9641 g = = 0. 92612 = 0. 9261 mass of sample 1. 0410 g mass of mercury 1. 4293 g = = 0. 92607 = 0. 9261 mass of sample 1. 5434 g mass of mercury 1. 1283 g = = 0. 92612 = 0. 9261 mass of sample 1. 2183 g Since each sample has the same proportion of mercury by mass, the data are consistent with the hypothesis that the material is a compound. 1. 103.