Title: Redox Reaction Introduction: Some of the most important chemical reactions are Redox reactions. They are also known as oxidation-reduction reactions. This reaction passes one or more electrons from one species to another. The species that loses electrons are being oxidized while the species that gains electrons are reduced. The reaction occurs simultaneously. Numbers are given to each element in a chemical reaction to help us find out which element is oxidized and which is
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|NaOCl + KI|4 |small white precipitate‚ color changed to black after starch was added | | with 1 drop of starch| || G.|KI + Pb(NO3)2|5 |turned bright yellow‚ milky texture| H.|NaOH + phenolphthalein|6 |turned dark pink‚ magenta| I.|HCl + phenolphthalein|7 |clear| J.|NaOH + AgNO3|8 |turned brownish green‚ cloudy| K.|AgNO3 + NH3|9 |turned light brown| |after exposure to bright light| || L.|NH3 and CuSO4|10 |light blue‚ cloudy| Part F: Testing of Various Foods with the NaOCl + KI solution
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down starch in solutions that are increasing in neutral pH. To do this the experiment was carried out so that tubes containing a reaction solution of the Amylase enzyme and starch were simultaneously mixed. The reactions were then introduced to I2-KI‚ which stopped the reactions‚ at two minute intervals. Each of these trials was repeated three times to ensure proper accuracy. After concluding the reactions they were placed into a spectrophotometer (A580) for analysis. Graphing the values of the
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Break Up Of Marks For Formative Assessment 3 & 4 For Class IX ENGLISH |1. |Monday Test |10 Marks | |2. |Literature based Role Play[ Bishop’s Candlestick] |3 Marks | |3. |MCB based Debate[content‚ Language‚ Pronunciation] |3 Marks | |
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the solution. The tip of the buret was checked to make sure there weren’t any bubbles in it. Then a 250mL beaker was obtained. A 25mL pipette was used to add exactly 25mL of the KIO₃ solution. Then 50mL of deionized water and about .25g of solid KI was added. The solution was stirred until the solid was completely dissolved. 2mL of glacial acetic acid was then added.
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bhi unnati hum‚ Waise to hai internet pe har prakaar ka gyaan‚ Par ache bure ki nahi hai use pehchaan . Nahin hain shabd kaise karoon dhanyavaad‚ Bas chahiye har pal aap sabka aashirvaad‚ Hoon jahan aaj mai usme hain bada yogdaan‚ Aap sabka jinhone diya mujhe itna gyaan. Aapne banaya hai mujhe is yogya‚ Ki praapt karoon mai apna lakshya‚ Diya hai har samay aapne sahaara ‚ Jab bhi laga mujhe ki mai haara. Par main hoon kitna matlabi‚ Yaad kiya na maine aapko kabhi‚ Aaj
Free Education Teacher Rajendra Prasad
SHORT HISTORY OF EKITI The word “Ekiti” as we know it today denotes a “mound.” and is derived from the rugged mountainous features of the State. It is an extensive State originally divided into 16 districts. Each district has its own “Owa” or King of which four are supreme. 1. The Owore of Otun 2. The Ajero of Ijero 3. The Ewi of Ado 4. The Elekole of Ikole 5. Alara of Ara 6. Alaaye of Eron-Alaye 7. Olomuo of Omuo
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Szegény Gélyi János lovai című novella elemzése „Megérti a négy okos állat a parádét…” A környezet‚ a természet legapróbb részletei is zseniálisan alátámasztják‚ olykor megindokolják‚ vagy kiegészítik a cselekmény folyamát Mikszáth novelláiban. A büszke‚ boldog lovak kicicomázott parádéja Gélyi János érzésének metaforája. Ez a metafora – Gélyi érzéseinek és a lovaknak párhuzama – végigvonul az egész művön. Bokros és Tündér a két ló neve‚ jelzős szerkezetként értelmezve Vér Klárát‚ a
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COPYRIGHT IN CINEMATOGRAPH FILMS Presentation At Asian Academy of Film and Television October 6‚ 2004 ©2004 LEX ORBIS Part - I October 6‚ 2004 ©2004 LEX ORBIS 2 Contents Idea – Expression Dichotomy Meaning of Copyright What do you protect Meaning of the “work” Works in which Copyright Subsist What is a Cinematograph Film Who is the owner of Copyright Owner in Cinematograph Film It is complex to ascertain the copyright in Cinematograph Film October 6‚ 2004 ©2004 LEX ORBIS
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method. Aim: To produce 1.50 g of lead iodide‚ PbI2‚ assuming 100 % conversion‚ using 1.00 g solid potassium iodide‚ KI‚ and 0.1 M lead nitrate‚ Pb(NO3)2 solution of volume 30 cm3 and hence calculate percentage yield. Calculations: 2KI(s) + Pb(NO3)2( aq) —> PbI2 (s) + 2KNO3 (aq) Mass of PbI2 to be produced = 1.5 g Moles of PbI2 = = 0.003 Moles of KI = 0.003*2 = 0.006 Mass of KI = 0.006*166 = 1.00 g Moles Pb(NO3)2 = 0.003 Volume Pb(NO3)2 = = 0.03 dm3 = 30 cm3 Background Information: Reaction:
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