Chemistry Module 1 revision notes - salts and redox reactions
Chemistry Module 1 Salts - A salt is an ionic compound with the following features: The positive ion or cation in a salt in a salt is usually a metal ion or an ammonium ion NH4 The negative ion or anion in a salt is derived from an acid Formation of Salts: Salts can be produced by neutralising acids with: - Carbonates - Bases - Alkalis Salts from bases: Acids react with bases to form a salt and water Salts from Carbonates: Acids react with carbonates to form a salt, CO2 and water Salts from alkalis: Acids react with alkalis to form a salt and water Salts from metals: Salts can be formed from the reaction of reactive metals with acids. There are known as redox reactions. Ammonia salts and fertilises: - Ammonia salts are used as artificial fertilisers - Ammonia salts are formed when acids are neutralised by aqueous ammonia Water of crystallisation -Water of crystallisation refers to water molecules that form an essential part of the crystalline structure of a compound. Often the compound cannot be crystallised if water molecules are not present. -The empirical formula of a hydrated compound is written in a unique way: -The empirical formula of the compound is separated from the water of crystallisation by a dot. - The relative number of water molecules of crystallisation is shown after a dot. Oxidation number: in a chemical formula each atom has an
Aim To study the effect of concentration of iodide ion solution on the rate of iodide ion I- oxidation by peroxodisulphate ion S2O82- using iodine clock reaction
INVESTIGATING THE RELATIONSHIP BETWEEN CONCENTRATION OF REACTANTS AND RATE OF REACTION USING IODINE CLOCK REACTION Aim To study the effect of concentration of iodide ion solution on the rate of iodide ion I- oxidation by peroxodisulphate ion S2O82- using iodine clock reaction Introduction In an iodine clock reaction, two clear solutions would be added together in a common container where no apparent reaction takes place. After a short delay, at a sudden, the clear solution would turn into a blue-black solution. In this experiment, two clear solutions - potassium iodide, KI and sodium peroxodisulphate, Na2S2O8 - would be added together, with delaying additives, where a blue-black product solution would be observed. The aim of this experiment is to measure the different time taken to form the blue-black solution for different concentration of potassium iodide solution used. The different rates of reaction for each concentration can then be determined because they are the reciprocal of the times taken. The chemical reaction that takes place is a redox reaction where iodide ion is oxidized and peroxodisulphate ion is reduced. The full ionic equation for the reaction is represented by: I- (aq) + S2O82- (aq) › I2 (aq) + 2SO42- (aq) * all potassium and sodium ions are spectator ions. However, without any delaying mechanism, the formation of the
chemistry limewater experiment
Chemistry-AS-Assessed Practical (Skills P and A) The aim of this experiment is to determine the concentration of limewater, in g dm-3, as accurately as possible using hydrochloric acid, HCl. In the experiment the hydrochloric acid must be diluted and then a titration can be done to find the concentration of limewater. The hydrochloric acid must be diluted as its concentration is too high. The concentration of the hydrochloric acid is exactly 2.00 mol dm-3 but it has to be to a similar concentration a calcium hydroxide so the titration can be done. The first step is to dilute the HCl and then the second step is to perform titration of the calcium hydroxide. The equation: 2HCl + Ca(OH)2 › CaCl2 + 2H20 This equation is a neutralisation reaction and will be used for the titration so the HCl that will be added to the limewater, that contains calcium hydroxide, will be neutralised. Equipments Burette- For measuring HCl as it is accurate. It can measure to 0.05cm3. Conical flask- For solutions to react in. Pipette- For measuring limewater solution. This is also very accurate. Pipette controller- To control amount of solution going in and out of pipette. Clamp stand- To hold burette. Clamps- To make sure equipments are secure. Funnel- So chemicals can be poured safely without spilling. White tile- To see clearly when reaction takes place. Distilled water- To clean
The Chemistry oh Phosphorous
The Chemistry of Phosphorus Among the non-metals, on the right hand side of the periodic table, with Atomic Number 15, is the element Phosphorus. The word 'phosphorus' is derived from the Greek word 'phosphoros' meaning 'bringer of light' and its discovery was completely accidental. It has the electronic configuration 2,8,5 and has 5 electrons in its outer most shell, hence it being in group 5 of the periodic table, under Nitrogen. In 1669, German alchemist Hennig Brand was attempting to create the 'philosopher's stone', which was a supposedly magical substance that would turn metals into gold, by evaporating urine. One day, after boiling urine into a paste, heating the paste to a high temperature and passing the vapours through water, where he hoped to find gold, was a mysterious white waxy substance that seemed to glow in the dark, and burst into flames when in contact with air. Brand had discovered phosphorus, well was the first to record this discovery. In fact, several other chemists could have discovered it at a similar time, and we now know that the substance he found was actually ammonium sodium hydrogen phosphate: (NH4)NaHPO41. This method produced only about 60g of phosphorus and used 1100L of urine, so was not very efficient, and it was only later on that investigators found alternative methods of obtaining it. One of its current processes of production
Additivity of Heats of Reaction: Hesss Law Design and Data Collection
IB Chem 11 Laura Hu Partner: Rhona Yue Worked together with Zheting and Melissa Additivity of Heats of Reaction: Hess's Law Purpose: To confirm Additivity of Heats of Reaction: Hess's Law. Supplies: As in text. Procedure: As in text. Reactions: Reaction #1: NaOH(s) + H2O(l) Na+ + OH- + q1 q1 is the difference in Heat for reaction 1 in Joules NaOH(s) is about 2.0g, and H2O(l) is 100mL Reaction #2: 100mL of 0.50M HCl(l) + about 2.0g NaOH(s) NaCl + H2O + q2 q2 is the difference in Heat for reaction 2 in Joules Reaction #3: 50.0mL 1.0M HCl(l) + 50.0mL 1.0M NaOH(l) NaCl + H2O + q3 q3 is the difference in Heat for reaction 3 in Joules Data Collection Table #3: Results collected by Rhona and Laura Initial Volume Mass of NaOH [m] Initial Final Temperature [t2] Reaction Temperature [t1] measured ± 0.005 (g) ± 0.2 (°C) ± 0.2 (°C) 00 ± 0.5mL of 2.93 21 22 H2O 2 00 ± 0.5mL of 2.87 20.9 26.5 0.50M HCl 50.0 ± 0.5mL 3 of 1.0M HCl ; N/A 20.4 26.7 50.0 ± 0.5mL of 1.0M NaOH Observation Notes: *Since these data is collected by us (Laura and Rhona), we have some observation notes about reaction 1, 2 and 3. Reaction 1: Temperature change is really slow. Reaction 2: Temperature raises a lot, but not very fast. Reaction 3: Temperature increases very fast. It jumped all the way to 26°C in 3 seconds right after pouring 50.0mL of 1.0M NaOH
Investigation into the chemist Fritz Haber
Coursework.info =) Research Project Chemist of choice: Fritz Haber Fritz Haber was born in the town of Breslau, Germany in early December 1868. His family was one of the oldest families in town, he was the son of Siegfried Haber who was a well know merchant in the town. Fritz studied at St. Elizabeth classical school in the town of his birth where he conducted many chemical experiments from a young age. From 1886 to 1891 Fritz studied chemistry at the University of Heidelberg under the academic advice of Robert Bunsen who had invented the Bunsen burner and also had help discover the elements cesium and rubidium. He also studied at the University of Berlin from guidance from A.W Hoffmann, and at the Technical School at Charlottenburg under Carl Liebbermann. Once he completed his studies at the universities he went and worked voluntarily for his father's chemical business as he was interested in chemical technology. He then went on to work at the under the eye off Professor Georg Lunge at the Institute of Technology in Zurich. After all this work he finally decided that he wanted to take up a scientific career and went to work with Ludwig Knorr at Hena for one and a half years to publish with him a joint paper on diacetosuccinic ester yet orthodox methods at the institute under Jena gave Haber little satisfaction. Still uncertain whether to devote himself to chemistry or
Decomposition of CuCO3
DECOMPOSITION OF CuCO? BACKGROUND THEORY Cu2O is a red crystalline material, which is produce by the electrolytic or furnace method. CuO is a black powder prepared by the ignition of suitable salts such as carbonate (1) The mole is the mass of substance that has the same number of particles as atoms in exactly 12g of carbon-12. One mole of any substance contains Avogadro's number 6.02×10²³ mol¯1 (2). Avogadro discovered that at room temperature (25°C) and pressure (1 atm), all gases occupy the same volume. 1 mole of any gas will occupy a volume of 24dm³. CALCULATING THE NEEDED MASS The best method is to take one of the two equations and try to prove or disprove it. Since equation 2 has only 1 mole of every substance this would be the easiest to work out in terms of ratio. This assumption is made because the equation is "stoichiometric" (3) meaning 1 mole of CuCO? decomposes to exactly 1 mole of CuO and 1 mole of carbon dioxide. However, 24000cm³ would be too much gas to produce in a school lab, as that size of apparatus is not available. Gas syringe will be used but has a maximum of 100cm³, so it would not be sensible to produce 100cm³ of gas because if equation 1 turns out to be correct (which would mean oxygen will also be produced), and there would consequently be more gas. Aiming to produce 75cm³ of gas would leave room for error and the possibility of
Investigation to determine the Relative atomic mass of Li
Investigation to determine the Relative atomic mass of Li Results and calculations for the first method. Mass of Li used : 0.12g Start volume of Water : 55.5ml Final volume of water : 250.0ml Volume of hydrogen produced : 194.5ml Now based on these results I t is possible to calculate the concentration of the limewater. Calculate the number of moles of hydrogen produced: We have 195ml of hydrogen that has been produced. - This is equal to 0.195 dm3 Since 24dm3 is one mole of hydrogen at room temperature. - 0.195 dm3 = 0.008125 mol Calculate the number of moles of Li reacted: Using the equation: 2Li (s) + 2H2O (l) --> H2 (g) + 2LiOH (aq) 2 : 1 One can see the reacting ratio is 2:1, the no. moles of Li will be twice that of H2. - Moles of Li = 2 x 0.008125 mol - Moles of Li = 0.01625 mol Calculate the relative atomic mass of Li: As the mass of Li and No of moles used are know, it is possible to calculate the relative atomic mass of Li. - Mass of Li = 0.12g - No Moles of Li = 0.01625 mol - Relative atomic mass = mass/no moles - 0.12/0.01625 = 7.3846 - Relative atomic mass of Li = 7.38 Results and calculations for the second titration method. 2 3 End Burette vol (ml) 46.70 43.95 43.70 Start Burette vol (ml) 06.50 01.65 01.30 Amount used (ml) 40.20 42.30 42.40 Using the two best titres (within 0.1 ml),
Titration Coursework
Titration Coursework Aim: My aim of this investigation is to find out the concentration of Sulphuric Acid (H2SO4) in a solution between 0.05 and 0.15 mol dm-3. During this investigation, I will be making a standard solution of Sodium Carbonate, of strength of 0.1 molar (M). In order to create this solution, I will be using a 250ml volumetric flask Firstly, I will need to know how much sodium carbonate I will need to use in my standard solution, which means that I will need to know the amount of moles and grams needed. For 1dm3: M = 0.1x106 = 10.6g For 250ml: 10.6 x 0.25 = 2.65g of Sodium carbonate By looking at above, you can see that by working out the amount of mass needed, I will need to use 2.65g of sodium carbonate to make my standard solution. Indicator solution For my solutions, I will be using an indicator solution, which is called methyl orange. This indicator is often chosen to be used in titrations because of its clear colour change. For the reason that it changes colour at the pH of a mid-strength acid, it is usually used in titrations for acids. Unlike a universal indicator, methyl orange does not have a full spectrum of colour change, but has a sharper end point. The reason why I will be using this indicator is because I am going to be using a strong acid and a weak alkali. As a result, for a titration that involves a strong acid and a weak alkali I
analysis of two commercial brands of bleaching solution
F.6A Lam Pik Sum (10) Title: analysis of two commercial brands of bleaching solution Date: 6/10/2008 Objective: to find out which of the two brands of bleach is cheaper base on their actual bleaching strength. Introduction Redox titration can be divided into two parts: in one half the reducing agent loses electrons and in other half the oxidizing agent gains electrons. The stoichiometric equation for a redox reaction may then derive using the fact that all the electrons lost by the reducing agent must be gained by the oxidizing agent. In this experiment, we need to find out the cost of active ingredient per gram in two different brands of household bleaches in other to find out with bleach is cheaper. The two bleaches we use are KAO ($13.9 for 1.5L) and Best Buy ($11.9 for 2L). Sodium hypochlorite forms the basis of most of commercial bleaches. In this analysis, the sodium hypochlorite is allowed to react with an excess of potassium iodide solution in the presence of acid, liberating iodine, which is then titrated against standard sodium thiosulphate solution. Reactions involved: ClO- + 2 I- + 2 H+ --> I2 + H2O + Cl- I2 + 2 S2O32- --> 2 I- + S4O62- In this redox titration, potassium iodide in the acidic medium acts as a reducing agent is added to the bleach solution to generate the iodine by the reduction of the hypochlorite ions. The formed iodine is then