628 lines
29 KiB
Plaintext
628 lines
29 KiB
Plaintext
ÜÜÜÜÜÜÜÜÜÜÜÜÜ ÜÜÜ ÜÜÜÜ
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ÜÛÛÛÛÛÛÛÛßÛßßßßßÛÛÜ ÜÜßßßßÜÜÜÜ ÜÛÜ ÜÛÛÛÛÛÛÛÛÜÜÜÜÜÛßß ßÛÛ
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ßÛÛÛÛÛÛÛÛÛÛÛÛÛÛÜ ßÛÛ ÜÛÛÛÜÛÛÜÜÜ ßÛÛÛÛÜ ßÛÛÛÛÛÛÛÜÛÛÜÜÜÛÛÝ Ûß
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ßßßÛÛÛÛÛÛÛÛÛÛÜ ÞÝ ÛÛÛÛÛÛÛÛÛÛÛßßÛÜÞÛÛÛ ÛÛÛÛÛÜ ßßÛÛÛÞß
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Mo.iMP ÜÛÛÜ ßÛÛÛÛÛÛÛÝÛ ÞÛÛÛÛÛÛÛÛÛ ÞÛÛÛÛ ÞÛÛÛÛÛÝ ßÛß
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ÜÛÛÛÛÛÛÛ ÛÛÛÛÛÛÛÛÝ ÞÛÛÛÛÛÛÛÛÝ ÛÛÛ ÛÛÛÛÛÛ
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ÜÛÛÛÛÛÛÛÝ ÞÛÛÛÛÛÛÛÛ ÞÛÛÛÛÛÛÛÛ ß ÞÛÛÛÛÛÛÜ ÜÛ
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ÜÛÛÛÛÛÛÛÝ ÛÛÛÛÛÛÛÛ ÛÛÛÛÛÛÛÛÝ ÞÞÛÛÛÛÛÛÛÛÛß
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ÜÛßÛÛÛÛÛÛ ÜÜ ÛÛÛÛÛÛÛÛÝ ÛÛÞÛÛÛÛÛÝ ÞÛÛÛÛÛÛßß
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ÜÛßÛÛÛÛÛÛÜÛÛÛÛÜÞÛÛÛÛÛÛÛÛ ÞÛ ßÛÛÛÛÛ Ü ÛÝÛÛÛÛÛ Ü
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ÜÛ ÞÛÛÛÛÛÛÛÛÛÛß ÛÛÛÛÛÛÛÛÛ ßÛÜ ßÛÛÛÜÜ ÜÜÛÛÛß ÞÛ ÞÛÛÛÝ ÜÜÛÛ
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ÛÛ ÛÛÛÛÛÛÛÛß ÛÛÛÛÛÛÛÛÛÛÜ ßÛÜ ßßÛÛÛÛÛÛÛÛÛß ÜÜÜß ÛÛÛÛÜÜÜÜÜÜÜÛÛÛÛÛß
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ßÛÜ ÜÛÛÛß ßÛÛÛÛÛÛÛÛÛÛÜ ßßÜÜ ßßÜÛÛßß ßÛÛÜ ßßßÛßÛÛÛÛÛÛÛßß
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ßßßßß ßßÛÛß ßßßßß ßßßßßßßßßßßßß
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ARRoGANT CoURiERS WiTH ESSaYS
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Grade Level: Type of Work Subject/Topic is on:
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[ ]6-8 [ ]Class Notes [A report on the causes/ ]
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[ ]9-10 [ ]Cliff Notes [effects/and other issues]
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[x]11-12 [x]Essay/Report [involvong acid rain. ]
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[ ]College [ ]Misc [ ]
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Dizzed: 07/94 # of Words:2,500 School: co-ed/public State: NY
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ÄÄÄÄÄÄÄÄÄ>ÄÄÄÄÄÄÄÄÄ>ÄÄÄÄÄÄÄÄÄ>Chop Here>ÄÄÄÄÄÄÄÄÄ>ÄÄÄÄÄÄÄÄÄ>ÄÄÄÄÄÄÄÄÄ>ÄÄÄÄÄÄÄÄÄ
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Introduction
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Acid rain has become an environmental concern of global importance
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within the last decade. With the increasing environmental awareness of the
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"unhealthy" condition of our planet earth the concern about acid rain has
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not lessened.
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In brief, acid rain is rain with pH values of less than 5.6. When
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dealing with acid rain one must study and understand the process of making
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Sulfuric acid. In this project we will take an in depth look into the
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production of sulfuric acid, some of its uses and the effects of it as a
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pollutant in our environment.
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Sulfuric Acid Industry in Ontario
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Among the many plants in Ontario where sulfuric acid is produced, there
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are three major plant locations that should be noted on account of their
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greater size. These are: (1) Inco. - Sudbury, (2) Noranda Mines Ltd. -
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Welland, and (3) Sulfide - Ontario
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There are a number of factors which govern the location of each
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manufacturing plant. Some of these factors that have to be considered when
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deciding the location of a Sulfuric Acid plant are:
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a. Whether there is ready access to raw materials;
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b. Whether the location is close to major transportation routes;
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c. Whether there is a suitable work force in the area for
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plant construction and operation;
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d. Whether there is sufficient energy resources readily
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available;
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e. Whether or not the chemical plant can carry out its
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operation without any unacceptable damage to the
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environment.
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Listed above are the basic deciding factors that govern the location of
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a plant. The following will explain in greater detail why these factors
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should be considered.
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1) Raw Materials The plant needs to be close to the raw materials that are
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involved in the production of sulfuric acid such as sulfur, lead,
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copper, zinc sulfides, etc..
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2) Transportation A manufacturer must consider proximity to transpor-
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tation routes and the location of both the source of raw materials and
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the market for the product. The raw materials have to be transported
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to the plant, and the final product must be transported to the customer
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or distributor. Economic pros and cons must also be thought about.
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For example, must sulfuric plants are located near the market because
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it costs more to transport sulfuric acid than the main raw materials,
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sulfur. Elaborate commission proof container are required for the
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transportation of sulfuric acid while sulfur can be much more easily
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transported by truck or railway car.
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3) Human Resources For a sulfuric acid plant to operate, a large work
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force will obviously be required. The plant must employ chemists,
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technicians, administrators, computer operators, and people in sales
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and marketing. A large number of workers will also be required for the
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daily operation of the plant. A work force of this diversity is
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therefore likely to be found only near major centres of population.
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4) Energy Demands Large amounts of energy will also be required for the
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production of many industrial chemicals. Thus, proximity to a
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plentiful supply of energy is often a determining factor in deciding
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the plant's location. 5) Environmental Concerns Most importantly,
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however, concerns about the environment must be carefully taken into
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consideration. The chemical reaction of changing sulfur and other
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substances to sulfuric acid results in the formation of other
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substances like sulfur dioxide. This causes acid rain. Therefore,
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there is a big problem about sulfuric plants causing damage to our
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environment as the plant is a source of sulfur emission leading to that
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of acid rain.
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6) Water Supplies Still another factor is the closeness of the location of
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the plants to water supplies as many manufacturing plants use water for
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cooling purposes. In addition to these factors, these questions must
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also be answered: Is land available near the proposed site at a
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reasonable cost? Is the climate of the area suitable? Are the general
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living conditions in the area suitable for the people involved who will
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be relocating in the area? Is there any suggestions offered by
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governments to locate in a particular region?
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The final decision on where the sulfuric acid plant really involves a
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careful examination and a compromise among all of the factors that have
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been discussed above.
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Producing Sulfuric Acid
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Sulfuric acid is produced by two principal processes-- the chamber
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process and the contact process.
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The contact process is the current process being used to produce
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sulfuric acid. In the contact process, a purified dry gas mixture
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containing 7-10% sulfur dioxide and 11-14% oxygen is passed through a
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preheater to a steel reactor containing a platinum or vanadium peroxide
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catalyst. The catalyst promotes the oxidation of sulfur dioxide to
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trioxide. This then reacts with water to produce sulfuric acid. In
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practice, sulfur trioxide reacts not with pure water but with recycled
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sulfuric acid.The reactions are:
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2SO2 + O2 ÄÄ> 2SO3
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SO3 + H2O ÄÄ> H2SO4
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The product of the contact plants is 98-100% acid. This can either be
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diluted to lower concentrations or made stronger with sulfur trioxide to
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yield oleums. For the process, the sources of sulfur dioxide may be
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produced from pure sulfur, from pyrite, recovered from smelter operations
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or by oxidation of hydrogen sulfide recovered from the purification of
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water gas, refinery gas, natural gas and other fuels.
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Battery Acid Industry
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Many industries depend on sulfuric acid. Among these industries is the
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battery acid industry.
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The electric battery or cell produces power by means of a chemical
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reaction. A battery can be primary or secondary. All batteries, primary or
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secondary, work as a result of a chemical reaction. This reaction produces
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an electric current because the atoms of which chemical elements are made,
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are held together by electrical forces when they react to form compounds.
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A battery cell consists of three basic parts; a positively charged
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electrode, called the cathode, a negatively charged electrode, called the
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anode, and a chemical substance, called an electrolyte, in which the
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electrodes are immersed. In either a wet or dry cell, sufficient liquid
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must be present to allow the chemical reactions to take place.
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Electricity is generated in cells because when any of these chemical
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substances is dissolved in water , its molecules break up and become
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electrically charged ions. Sulfuric acid is a good example. Sulfuric acid,
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H2SO4, has molecules of which consist of two atoms of hydrogen, one of
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sulfur and four oxygen. When dissolved in water, the molecules split into
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three parts, the two atoms of hydrogen separate and in the process each
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loses an electron, becoming a positively charged ion (H+). The sulfur atom
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and the four atoms of oxygen remain together as a sulfate group (SO4), and
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acquire the two electrons lost by the hydrogen atoms, thus becoming
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negatively charged (SO4--). These groups can combine with others of
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opposite charge to form other compounds.
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The lead-acid cell uses sulfuric acid as the electrolyte. The
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lead-acid storage battery is the most common secondary battery used today,
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and is typical of those used in automobiles. The following will describe
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both the charging and discharging phase of the lead-storage battery and how
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sulfuric acid, as the electrolyte, is used in the process. The lead
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storage battery consists of two electrodes or plates, which are made of
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lead and lead peroxide and are immersed in an electrolytic solution of
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sulfuric acid. The lead is the anode and the lead peroxide is the cathode.
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When the battery is used, both electrodes are converted to lead sulfate by
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the following process. At the sulfate ion that is present in the solution
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from the sulfuric acid. At the cathode, meanwhile, the lead peroxide
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accepts two electrons and releases the oxygen; lead oxide is formed first,
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and then lead joins the sulfate ion to form lead sulfate. At the same
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time, four hydrogen ions released from the acid join the oxygen released
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from the lead peroxide to form water. When all the sulfuric acid is used
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up, the battery is "discharged" produces no current. The battery can be
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recharged by passing the current through it in the opposite direction.
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This process reverses all the previous reactions and forms lead at the
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anode and lead peroxide at the cathode.
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Proposed Problem
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i) The concentration of sulfuric acid is 0.0443 mol/L.
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The pH is: No. mol of hydrogen ions = 0.0443 mol/L x 2
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= 0.0886 mol/L hydrogen ions
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pH = - log [H]
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= - log (0.0886)
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= - (-1.0525)
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= 1.05
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Therefore, pH is 1.05.
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ii) The amount of base needed to neutralize the lake water
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is:
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volume of lake = 2000m x 800m x 50m
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= 800,000,000 m3 or 8x108 m3
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since 1m3=1000L, therefore 8x1011 L
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0.0443 mol/L x 8x1011 = 3.54 x 1010 mol of H2SO4 in water
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# mol NaOH = 3.54 x 1010 mol H2SO4 x 2 mol NaOH
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1 mol H2SO4
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= 7.08 x 1010 mol of NaOH needed
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Mass of NaOH = 7.08 x 1010 mol NaOH x 40 g NaOH
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1 mol NaOH
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= 2.83 x 1012 g NaOH
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or 2.83 x 109 kg NaOH
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Therefore a total of 2.83 x 1012 g of NaOH is needed to
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neutralize the lake water.
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iii) The use of sodium hydroxide versus limestone to
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neutralize the lake water:
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Sodium hydroxide: Sodium hydroxide produces water when reacting with
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an acid, it also dissolves in water quite readily. When using sodium
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hydroxide to neutralize a lake, there may be several problems. One problem
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is that when sodium hydroxide dissolves in water, it gives off heat and
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this may harm aquatic living organisms. Besides this, vast amounts of
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sodium hydroxide is required to neutralize a lake therefore large amounts
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of this substance which is corrosive will have to be transported. This is
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a great risk to the environment if a spill was to occur.
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The following equation shows that water is produced when using sodium
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hydroxide.
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2NaOH + H2SO4 ÄÄ> Na2 SO4 + 2H2O
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Limestone: Another way to neutralize a lake is by liming. Liming of
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lakes must be done with considerable caution and with an awareness that the
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aquatic ecosystem will not be restored to its original pre-acidic state
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even though the pH of water may have returned to more normal levels. When
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limestone dissolves in water it produces carbon dioxide. This could be a
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problem since a higher content of carbon dioxide would mean a lowered
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oxygen content especially when much algae growth is present. As a result,
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fish and other organisms may suffer. Limestone also does not dissolve as
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readily as sodium hydroxide thus taking a longer period of time to react
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with sulfuric acid to neutralize the lake. The equation for the
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neutralization using limestone is as follows:
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Ca CO3 + H2SO4 ÄÄ> CaSO4 + H2O.
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iv) The effect of the Acid or excessive Base on the plant
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and animal life:
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You will probably find that there aren't many aquatic living organisms
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in waters that are excessively basic or acidic. A high acidic or basic
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content in lakes kill fishes and other aquatic species. Prolonged exposure
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to acidic or excessively basic conditions can lead to reproductive failure
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and morphological aberration of fish. A lowered pH tends to neutralize
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toxic metals. The accumulation of such metals in fish contaminates food
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chains of which we are a part as these metals can make fish unfit for human
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consumption. Acidification of a lake causes a reduction of the production
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of phytoplankton (which is a primary producer) as well as in the
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productivity of the growth of many other aquatic plants. In acidic
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conditions, zooplankton species will probably becompletely eliminated. In
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addition, bacterial decomposition of dead matter is seriously retarded in
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acidified lake waters. Other effects of acidic conditions arean
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overfertilization of algae and other microscopic plant lifecausing algae
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blooms. Overgrowth of these consumes quickly most of the oxygen in water
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thus causing other life forms to die from oxygen starvation.
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When there are excessive base or acid in waters, not only do aquatic
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organisms get affected but animals who depend on aquatic plants to survive
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will starve too, since few aquatic plants survive in such conditions.
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Therefore each organism in the aquatic ecosystem is effected by excessive
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basic or acidic conditions because anything affecting one organism will
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affect the food chain, sending repercussions throughout the entire
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ecosystem.
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v) The factors that govern this plant's location, if this plant employs
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40% of the towns people:
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The major factors that would govern this plant's location would be
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whether there is ready access to raw materials; whether the location is
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close to major transportation routes; whether energy resources are readily
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available and if there is an adequate water supply in the area. Since this
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plant would employ 40% of the towns people, the plant should be close to
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the town while still far enough so that in case of any leakage of the
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plant, the town will be within a safe distance of being severely affected.
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The factor of whether the general living conditions in the area are
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suitable for the workers should also be considered as well.
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Additional Comments
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a) The situation of pollution in the Great Lakes and process being used to
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start cleaning it up--comments: Everyday, roughly 3630 kilograms of
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toxic chemicals enter the lakes, nearby land and air. Pollution of the
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Great Lakes has become an increasingly serious problem. Just in Lake
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Ontario, hundreds of thousands of tons of contaminants have been
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deposited over the years. These include DDT, PCBs, mercury, dioxins
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and mirex, a pesticide. About 4.6 million people depend on Lake
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Ontario alone for drinking water. The environmental problem of
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greatest concern to Lake Ontario neighbours is water-discharged toxic
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chemicals and industrial air pollutants. Not only is this occurring in
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Lake Ontario but the other Great Lakes as well. The lakes probably
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have all these poisonous chemicals in them: salts drained from urban
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streets, coliform bacteria from the sewage civilization plus a
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selection of substances such as phosphorus, polychlorinated biphenyls
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and heavy metals. It is reported that the toxic chemicals in the Great
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Lakes basin are a health risk linked to brain damage, birth defects and
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cancer. All the predator species at the top of the food chain have
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shown health problems as a result of toxic chemicals building up in
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their bodies. Chemicals that exist in low levels in the air and water
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accumulate as they move up through the food chain. At present 35
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million humans who live around the horridly polluted five Great Lakes
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face increasing health risks from environmental contaminants. Millions
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of people in the Great Lakes are exposed to hazardous chemicals. They
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drink them in the contaminated water, eat them concentrated in the
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flesh of the fish and breathe them in the air.
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Mulroney said that the risks are too high and that we cannot afford any
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more risks. He said pollution problems could be fought under a three-stage
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plan over the next decade:
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1) A "toxic freeze" banning new polluters from putting
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up pipes or smokestacks in the region
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2) An attack on "non-point sources" of pollution, such
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as run-off from streets and farms where groundwater
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is loaded with pesticides.
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3) A crackdown on existing polluters when their smoke
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and sewer-discharge permits come up for renewal,
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requiring them to scale down their pollution.
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Consumers can also help by demanding pesticide-free food.
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International agreements have been made to clean up the Great Lakes.
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Canada's federal Conservative government has announced in 1989 to spend
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$125 million over five years on Great Lakes cleanup. By one estimate, it
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may cost as much as $100 billion to retrieve the purity of the Great Lakes
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once had.
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b) The treatment of water for drinking and water purifiers one can
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purchase--comments: As the people's uncertainty to the quality of our
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drinking water increases, many more people are buying water treatment
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devices and purifiers. Even though most treated tap water is fit to
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drink, people are losing faith in the government to keep it that way.
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therefore purifier leave become increasingly popular among consumers.
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However each of the most popular cleansing methods has some
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disadvantages. Many filters use some form of "activate" carbon.
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However, few carbon filters alone do a very good job of reducing heavy
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metals such as lead even though the smallest sink-tap charcoal strainer
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will make cloudy water look and taste a bit better. Distillation units
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turn water to steam and recondense it to a cleaner state. This process
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has its disadvantages, too for they can also pass along harmful
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chemicals with low boiling points into the water. Another water
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treatment device is the reverse-osmosis device which uses sophisticate
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membranes to separate pure water from impure. Even though this is
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effective, three gallons of water for every good one produced is
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generally wasted. Some machines zap germs with lethal doses of
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ultraviolet light. A specific example of a water filter is the NSA
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3000HM high density filter. This filtration unit is designed to remove
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lead, iron, sulfur and manganese from your drinking water supply. Still
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another example is a water treatment system called the NSA Bateriostatic
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water treatment system. This system removes chlorine, bad taste and
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odours, reduces undissolved particles (sediment, discolouration, etc.)
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and inhibits bacteria growth.
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Each of these processes can reduce impurities in your water supply and
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many machines as suggested by the above examples combine several
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approaches.
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c) BRIEF OUTLINE OF THE KEY EVENTS IN THE U.S.-CANADA
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RELATIONS WITH RESPECT TO CLEANING UP THE GREAT LAKES:
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1972: the U.S. chairman of the International Joint
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Commission, announced to study to determine the polluting
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effects on the Great Lakes urban development and agricultural
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land use, find remedies and estimate cleanup costs; Canada
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and the United States signed a Great Lakes Quality Agreement.
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1974: Canadians say the cleanup financed by Washington
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is already running far behind the schedule
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envisaged when the agreement was signed.
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1978: Canada and the United States agreed to the goal
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of zero discharge of pollution. 1987: the
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goal made in 1978 is made again, this means both
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countries agreed to work toward completely
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eliminating persistent toxic pollutants, not just
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the amount being discharged by industry; Mulroney
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also proposed that the U.S. slash industrial
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sulfide and nitrogen oxide emissions by half
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before 1994.
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The Canada-U.S. International Joint Commission meets every two years to
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discuss pollution and other issues concerning the Great Lakes, At present,
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they are making a ten-year headline for the Great Lakes to be cleaned up.
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Bibliography
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Encyclopedias
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Collier Encyclopedia, volume 3, U.S.A.: MacMillan
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Educational Company, New York, 1984.
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Encyclopedia of Industrial Chemical Analysis, volume 18,
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U.S.A.: John Wiley & Sons, Inc, 1973.
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Science & Technology Illustrated: The World Around U.S.,
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Volume 3, U.S.A: Encyclopedia Britannica Inc, 1984.
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Articles
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Cleaning Up By Cleaning Up Newsweek: Feb. 27, 1989.
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"Deadline Urged for Cleanup of Great Lakes", Toronto Star,
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Oct. 14, 1989.
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"Great Afflictions of the Great Lakes", The Globe and Mail,
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Oct. 14, 1989.
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"Great Lakes Pollution as a Political Issue", The Globe and
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Mail, Oct. 16, 1989.
|
|
|
|
"N.Y. Accused of Overlooking Pollution in Lake", Toronto
|
|
Star, Feb. 26, 1990.
|
|
|
|
"Pact On Great Lakes Cleanup Not Working, Greenpeace Says",
|
|
Globe and mail, July 19, 1989.
|
|
"The Clean Water Industry Grows on Fear, Uncertainty",
|
|
Toronto Star, Jan. 28, 1990.
|
|
|
|
"Information Scarce On Great Lakes Chemicals", The Globe and
|
|
Mail, Oct. 14, 1989.
|
|
|
|
Others
|
|
|
|
Countdown Acid Rain, Facts: Ministry of the Environment,
|
|
1989.
|
|
Sanderson, Kimberly, Acid Forming Emissions, Canada:
|
|
Environment Council of Edmonton, Alberta, 1984.
|
|
|
|
The New How It Works, volume 2, Westport Connecticut; H.S.
|
|
Stuttman Inc., 1987.
|
|
|
|
Weller, Phil., Acid Rain: Silent Crisis, Canada: Between the
|
|
Lines, 1980.
|
|
TITRATION LABORATORY
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Purpose: 1) to prepare 0.1 mol/L NaOH solution.
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2) to standardize the NaOH solution in part 1, using
|
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potassium hydrogen phthalate.
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3) to determine the unknown molarity of a H2SO4 solution using
|
|
standardized solution.
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Part 1 - Prepare 0.1 mol/L NaOH solution
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|
|
Observations:
|
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Data:
|
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mass of NaOH + paper tray = 4.58 g
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mass of paper tray = 3.46 g
|
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mass of NaOH pellets = 1.12 g
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Calculation:
|
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Number of mole of NaOH = mass of NaOH pellets = 1.12g = 0.028mol g.
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mol mass of NaOH 40g
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|
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Conclusion:
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Questions:
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1. When the NaOH pellets are left in the atmosphere, it reacts with
|
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the gases and absorbs water (moisture) in the air making it unable
|
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to neutralize too well.
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2. The gram mole mass of a substance is the mass in gram of 1 mol of
|
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that substance.
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3. The solution of NaOH must be standardized in order to accurately
|
|
calculate the concentration of the acid.
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Part 2 - Standardize the NaOH solution prepared in Part 1, using
|
|
potassium hydrogen phthlate
|
|
|
|
Observations:
|
|
Data:
|
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mass of vial + KpH = 22.19g
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mass of vial + KpH after
|
|
transfer to 1st flask = 22.19g
|
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mass of vial + KpH after
|
|
transfer to 2nd flask = 21.93g
|
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mass of vial + KpH after
|
|
|
|
ÚÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ¿
|
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³ flask ³ mass of KpH ³ volume of NaOH ³ conc. of NaOH ³
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ÃÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ´
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³ 1 ³ .12 ³ 1.2 mL ³ 0.00071 ³
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ÃÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ´
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³ 2 ³ .14 ³ 1.5 mL ³ 0.00103 ³
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ÀÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÙ
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|
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flask 1
|
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To calculate the concentration of NaOH (mol/L) the number of moles of
|
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KpH have to be calculated. No. of mol of KpH = 0.12
|
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204g/mol
|
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= 5.9 x 10-4 mol
|
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The ratio of KpH to NaOH is 1:1
|
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Therefore, the no. of NaOH = 5.9 x 10-4mol.
|
|
The equation being used is: KpH + NaOH --> KHC8H3NaO4+H2O
|
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The following equation is used to calculate the concentration of NaOH.
|
|
c = n n=number of mol = 5.9 x 10-4mol
|
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v v=volume = 1.2 x 10-3
|
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c = 5.9 x 10-4mol c=concentration = ?
|
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1.2 x 10-3L
|
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c = 0.492 mol/L
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|
|
Therefore, the NaOH solution in Flask 1 is 0.492 mol/L.
|
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flask 2
|
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No. of mol of KpH = 0.14
|
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204g/mol
|
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= 6.9 x 10-4 mol
|
|
The ratio of KpH to NaOH is 1:1
|
|
Therefore, the no. of NaOH = 6.9 x 10-4mol.
|
|
c = n n=number of mol = 6.9 x 10-4mol
|
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v v=volume = 1.5 x 10-3
|
|
c = 6.9 x 10-4mol c=concentration = ?
|
|
1.5 x 10-3L
|
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c = 0.46 mol/L
|
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|
|
Therefore, the NaOH solution in Flask 2 is 0.46 mol/L.
|
|
|
|
The average molarity of NaOH solution = flask 1 + flask 2
|
|
2
|
|
= 0.492 + 0.46 mol/L
|
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2
|
|
= 0.476 mol/L
|
|
|
|
Conclusions:
|
|
Questions:
|
|
1. The equation for the neutralization of potassium hydrogen phthalate
|
|
solution with NaOH solution is:
|
|
|
|
NaOH + KHC8H4O4 --> KHC8H3O4Na + H2O
|
|
|
|
2. The primary error in this titration process is that it is very easy
|
|
to go over the endpoint. We can improve this by being very careful
|
|
when letting the NaOH solution into the acidic solution. Especially
|
|
when we see that the pink colour is starting to stay we should
|
|
allow only part drops of the NaOH solution into the acidic solution
|
|
to make certain that we do not go over the endpoint.
|
|
|
|
3. The endpoint of a titration is the point at which the number of
|
|
moles of hydroxide ion added is the same as the number of moles of
|
|
hydrogen ion originally present in the flask. The difference
|
|
between the stoichiometric point and endpoint of a reaction is that
|
|
the stoichiometric point is exactly the point at which the number
|
|
of moles of hydroxide ion is equal to the number of moles of
|
|
hydrogen ion while the endpoint is usually a little over this point
|
|
when the solution has turned pink.
|
|
|
|
4. Phenolphthalein was chosen as the indicator of this titration
|
|
because phenolphthalein is a dye that is colourless in acidic
|
|
solutions but shows-up bright red or pink in basic solutions.
|
|
|
|
5. An indicator is a compound that detects the presence of acids and
|
|
bases by changing to different colours.
|
|
|
|
Part 3 - To determine the unknown molarity of a H2SO4 solution using
|
|
standardized NaOH solution.
|
|
|
|
Observations:
|
|
|
|
ÉÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÑÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÑÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ»
|
|
º Volume of known ³ Volume of known ³ Molarity of º
|
|
ºsurfuric acid sol'n(mL)³ conc. of NaOH(mL) ³ sulfuric acid º
|
|
ÌÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍØÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍØÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ͹
|
|
º 1. 25 mL ³ 4.5 mL (0.0045L) ³ 3.915 x 10-6 º
|
|
ÇÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄĶ
|
|
º 2. 25 mL ³ 4.8 mL (0.0048L) ³ 4.176 x 10-6 º
|
|
ÈÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÏÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÏÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍͼ
|
|
To find the molarity of unknown sulfuric acid solution:
|
|
Equation of reaction:
|
|
2NaOH + H2SO4 --> Na2SO4 + 2H2O
|
|
|
|
General equation solve:
|
|
Ca Va = Cb Vb a = acid
|
|
na nb b = base
|
|
|
|
|
|
Flask 1
|
|
For NaOH (base):
|
|
v = 4.5 x 10-3 L c = 0.476 mol/L n = ?
|
|
n = v c
|
|
= 4.5 x 10-3 x 0.476
|
|
= 2.1 x 10-3 mol NaOH
|
|
|
|
For H2SO4 (acid):
|
|
v = 0.025 L
|
|
n = ?
|
|
c = ?
|
|
#mol of H2SO4 = 2.1 x 10-3 x 1 H2SO4 2
|
|
NaOH
|
|
= 1.05 x 10-3
|
|
|
|
Solving the equation:
|
|
Ca Va = Cb Vb
|
|
na nb
|
|
|
|
Ca x 0.025 L = 0.476 mol/L x 0.0045 L
|
|
1.05 x 10-3 2.1 x 10-3 mol
|
|
|
|
Ca = 0.476 x 0.0045 x 1.05 x 10-3
|
|
2.1 x 10-3 x 0.025
|
|
|
|
Ca = 2.25 x 10-6
|
|
5.25 x 10-5
|
|
Ca = 0.0429 mol/L
|
|
|
|
Flask 2
|
|
Ca x 0.025 L = 0.476 mol/L x 0.0048 L
|
|
1.14 x 10-3 2.28 x 10-3
|
|
|
|
Ca = 0.476 x 0.0048 x 1.14 x 10-3
|
|
2.28 x 10-3 x 0.025
|
|
|
|
Ca = 0.0457 mol/L
|
|
|
|
The average molarity of H2SO4 solution = flask 1 + flask 2
|
|
2
|
|
= 0.0429 + 0.0457
|
|
2
|
|
= 0.0443 mol/L
|