Monoatomic Ions Name Symbol Name hydrogen ion H¯ hydride lithium ion F¯ fluoride Note that the sodium ion Cl¯ chloride letters in an ion’s potassium ion Br¯ bromide name before the rubidium ion I¯ iodide -ide ending is 2¯ cesium ion O oxide the stem. For 2¯ beryllium ion S sulfide example‚ the stem magnesium ion Se2¯ selenide for bromide is calcium ion Te2¯ telluride brom-. strontium ion barium ion Ag + silver ion N 3¯ nitride radium ion Ni2+ nickel
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or loses electrons while being bonded with another atom an ion is formed. This bond causes an atom to become either a positive or negative ion. Electrons have a negative charge‚ meaning that if an atom loses an electron‚ the amount of protons are greater than electrons. This makes the atom turn into a positive ion which is known as a cation. The opposite of this is known as an anion‚ which is when an atom gains electrons and becomes an ion that is negative. Non-metals form anions and metals form cations
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called ions. Those ionic substances are broken down by electrolysis. Electrolysis is the breakdown of a substance by electricity‚ and it only happens in liquids. The liquids that can be electrolyzed are called electrolytes. When said electrolyzed‚ means that the compounds in solutions are broken down when they conduct electricity. Its necessary the use of electrodes since it conducts the electricity to the solution. The electrodes are needed to be an anode‚ which receives the negative ions‚ and the
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The first step in both mechanisms is the protonation of the alcohol to form an oxonium ion‚ converting the OH group into a good leaving group. What happens next depends on the nature of the alkyl group‚ R. If R is a group that readily forms a carbocation‚ then the slow‚ rate-determining step is the loss of a water molecule from the oxonium ion. Once formed‚ the carbocation then reacts rapidly with a halide ion to form the alkyl halide. SN1 Mechanism: The first step is protonation of the alcohol‚
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VIHS/ Department of Chemistry Chemistry Revision Unit II (Edexcel) (01) a) When lithium nitrate and sodium nitrate are heated separately‚ both decompose giving oxygen gas as one of the products. (i) Which of these two nitrates would decompose at the lower temperature? .........................................................………………………………………….................... (ii) Give the name of any other product formed when sodium nitrate is heated. .....................................................
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reactions below‚ (i) state and explain what would be observed; and (ii) write an appropriate equation. (a) adding excess sodium sulphite solution to iodine solution (b) adding aqueous chlorine to potassium bromide solution (c) adding excess potassium iodide solution to acidified potassium permanganate solution 1 (d) adding excess iron(II) sulphate solution to acidified potassium dichromate solution (e) adding concentrated nitric acid to magnesium ribbons (f) (g) (h) (i) (j) (k) (l) (m) (n) 5. adding
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CHEMICAL KINETICS: IODINE-CLOCK REACTION DATE SUBMITTED: 14 DECEMBER 2012 DATE PERFORMED: 7 DECEMBER 2012 ABSTRACT Chemical kinetics involving reaction rates and mechanisms is an essential part of our daily life in the modern world. It helps us understand whether particular reactions are favorable and how to save time or prolong time during each reaction. Experiment demonstrated the how concentration‚ temperature and presence of a catalyst can change the rate of a reaction. 5 runs of dilution
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HKDSE CHEMISTRY – A Modern View (Chemistry) Experiment Workbook 5 Suggested answers Chapter 52 Importance of industrial processes Chapter 53 Rate equation Experiment 53.1 Determining the rate equation of a reaction using method of initial rate (A microscale experiment) 1 Chapter 54 Activation energy Experiment 54.1 Determining the activation energy of a chemical reaction 3 Chapter 55 Catalysis and industrial processes Experiment 55.1 Investigating the action of a catalyst 6 Experiment
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The reaction of Hydrogen Peroxide and Iodide ions in an acidic medium: 3 I-(aq) + H2O2(aq) + 2 H+(aq) I3-(aq) + 2 H2O(l) Step 1. H2O2(aq) + I-(aq) IO-(aq) + H2O(l) Step 2. IO-(aq) + H+(aq) HOI(aq) Step 3. HOI(aq) + 2 I-(aq) + H+(aq) I3-(aq) + H2O(l) In this reaction the three iodide ions are oxidised to form the triiodide ion. This occurs in three steps. Firstly‚ the peroxide molecule oxidises a single iodide ion‚ to form a hypoiodite ion‚ and a molecule of water. This is the
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the reducing solution is potassium iodate solution and the oxidizing solution is sodium thiosulphate solution. Potassium iodate solution which is an oxidizing agent is added into an excess solution of acidified potassium iodide. This reaction will release iodine. Potassium iodide is acidified with sulphuric acid and the iodine released quickly titrated with sodium thiosulphate until it become light yellow. The iodine then detected with starch solution and it turn into dark blue solution and titrated
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