To determine the concentration of limewater
To determine the concentration of limewater The aim of my experiment is to find the concentration of limewater solution provided. To do this I am provided with the following chemicals. Limewater: 250cm (1g dm ) Hydrochloric acid at concentration of 2 mol dm As you can see the hydrochloric acid is too concentrated for titration (i.e. one drop could change the colour of the solution.) therefore this acid will need to be diluted. Background knowledge: Limewater can be made by dissolved slacked lime in water to get a solution of calcium hydroxide (Ca (OH) 2 ) Quicklime + water --------> slaked lime CaO (s) + H2O (l) --------> CaCl2 (aq) + 2H2O (l) In this reaction it shows the calcium oxide reacts with water to produce calcium hydroxide. After adding more water to calcium hydroxide, it produces the saturated aqueous solution known as limewater. My experiment will be based on this neutralisation reaction between the limewater and hydrochloric acid. Ca(OH) 2 (aq) + HCl (aq) --------> CaCl2 (aq) + 2H2O (l) (back ground knowledge of limewater provided by AS chemistry 1 page...) Diluting hydrochloric acid: My first task for this experiment is to dilute the hydrochloric acid. To do this I first needed to work out how many moles I wanted for hydrochloric acid. For this part of my calculation I choose to use text book Calculations for chemistry to help me.
How oil is obtained, used and effects our world.
How are products from oil obtained and used? Oil. What actually is oil? Oil is or was originally made of organic material, like plankton. When the plankton dies it sinks to the sea floor. Over time it becomes part of the many layers of sedimentary rock that form there. Lots and lots of layers build up putting pressure on the lower layers, turning them into petroleum. If there is enough heat and pressure it will turn into crude oil. Crude oil is made up of different carbon-based fractions. These can be extracted using a fractionising column. All these fractions have different boiling points, this means that when heated, certain parts of the crude oil will evaporate and can therefore be collected by condensing it. This means that if you know the certain boiling points of each fraction you can separate the certain fraction you need, which is used for different products. This is known as fractional distillation. This shows the different heat at which some of the fractions boil at. As you can see the boiling points range from 20? to 400?. The different boiling points are important because this means that you can separate out the crude oil for the desired separate fraction. Some of the fractions are; Kerosene- a liquid hydrocarbon usually used to power aircraft or for heating. Naphtha- this is used to create a high octane gasoline. The higher the octane rating, the more energy is
Investigation into the effect of acid/alkali strength on the heat change when acids and alkalis are mixed
Investigation into the effect of acid/alkali strength on the heat change when acids and alkalis are mixed Planning The main aim of this experiment is to investigate the heat change when acids and alkalis are mixed. The temperature of the acid and alkali that are going to be mixed will be taken prior to mixing, and after they have been mixed in a polystyrene cup. The maximum temperature rise will be noted as this will be the biggest heat change that has occurred. The highest temperature after mixing and the temperature prior to mixing will be subtracted to give the heat change. To ensure that this is a valid test the volume of the acid and alkali will be kept constant at 40cm3. The volume will be kept constant because if there were a differing volume of acid to alkali this would have an influence on the temperature rise as there is not the same amount of solution. The only variable in this experiment will be the strength of the acid and alkali. This will allow us to examine the manner in which the heat evolved differs for differing strengths. Comparisons can then be undertaken to see how concentration affects the heat change in set volumes of acid and alkalis. When the acid and alkali of the same volume are mixed, this will cause the process of neutralisation to occur. Neutralisation is the reaction between an acid and a base. It is the formation of a bond between
To determine the amount of ammonia in a sample of household cleaning product, 'cloudy ammonia', in the form of NH4OH through the process of volumetric analysis.
Volumetric Analysis of Ammonia in Household Cleaning Product Aim: To determine the amount of ammonia in a sample of household cleaning product, 'cloudy ammonia', in the form of NH4OH through the process of volumetric analysis. Introduction: Neutralisation refers to the process whereby an acid reacts with a base in stoichiometric proportions to each other to form a salt and water. In this experiment, the neutralisation reaction can be summarized by the following equation: HCl (aq) + NH4OH (aq) NH4Cl (aq) + H2O (l) In this prac, the primary standard is Na2CO3. Primary standards are substances that possess certain properties (i.e. it is soluble), which enable it to be made up into a solution of a known concentration with distilled water to high degree of accuracy. A secondary standard on the other hand, is any solution that has an accurately known concentration. In this experiment, HCl acid is the secondary standard. The standardising of HCl can be summarized by the following equation: 2HCl (aq) + Na2CO3 (aq) 2NaCl(aq) + CO2 (g) + H2O (l) Since sodium carbonate is deliquescent, it is kept in a desiccator to prevent it from absorbing water vapour from the atmosphere. The equivalence point of a titration is the point where the reactants are present in stoichiometric proportions to each other. The end point of a titration is the stage at which the chosen
Investigate the effect of changing the concentration of sodium hydroxide (alkali) on the volume of hydrochloric acid needed to neutralize a fixed volume of alkali by measuring the temperature and noting colour changes of the solution mixture.
College: Brookhouse six form Student: Miss Vincia Phillip Teacher: Mrs Zainab Topic: Titration/ neutralisation reactions Title: Titration/ Neutralisation Aim: To investigate the effect of changing the concentration of sodium hydroxide (alkali) on the volume of hydrochloric acid needed to neutralize a fixed volume of alkali by measuring the temperature and noting colour changes of the solution mixture. Background knowledge: Neutralization occurs when an acid is made to react with a base. In this reaction, the chemical opposites cancel each other out. Alkalis contain OH? (aq) ions, whereas acids contain H? (aq) ions. In the neutralization reaction the OH? and H? ions come together to form water; (H O), which is a neutral substance. The reaction to be carried out will be between Sodium hydroxide (NaOh) and Hydrochloric acid (HCl). Equations: Sodium hydroxide +hydrochloric acid›Sodiumchloride+ Water NaOh (aq) +HCl (aq) ›NaCl (aq) +H O (aq) Equation: H? (aq) + OH? (aq) ›H O (l) In this reaction there are four different ions. These are Na, OH?, H and Cl?. The reactivity series shows elements in order of their reactivity. Reactivity is based on how vigorously elements react with oxygen, air and dilute acid. In displacement reactions, more reactive elements displace others from their compounds and take their place. Spectator ions are ions which are indirectly
Biology Practical Investigation to find the lowest concentration of Copper Sulphate solution that will denature egg albumen.
Biology Practical Investigation to find the lowest concentration of Copper Sulphate solution that will denature egg albumen Introduction Proteins are organic compounds of large molecular mass. They are not truly soluble in water; instead they form colloidal suspensions. Proteins are made up of amino acids.. There are three main types of bonding. The first is a disulphide bond. This occurs between cysteine molecules in the same amino acid chain (intrachain) or between molecules in different chains (interchain). This is the strongest and last bond to break. The second bond is an ionic bond. At a suitable pH, an interaction may occur between ionised amino and carboxylic groups. This forms an ionic bond. These ionic bonds are weak and can be broken in an aqueous solution by changing the pH of the medium surrounding the polypeptide. The third bond is the hydrogen bond, which occurs between hydrogen and oxygen atoms within the polypeptide chain. Hydrogen bonds are not strong on their own, although a large number of them together makes them very strong. The primary structure is the linear sequence of amino acids stablised by peptide bonds. There are an endless number of different possible primary structure The secondary structure is the sequence of amino acids arranged as either an alpha helix of a beta pleated sheet stabilised by peptide s. bonds and hydrogen bonds. In an
Formula of a hydrated salt
Determination of the formula of hydrated iron (II) sulphate Analysis Method 1: . A crucible was weighed to two decimal places using an accurate balance. 2. Approximately 1 g of hydrated iron (II) sulphate was added to the crucible, which was then re-weighed. 3. The crucible was heated gently for about two minutes in a fume cupboard. 4. The crucible was allowed to cool before weighing it again. 5. Steps 3 & 4 were repeated. Results and Calculations: Mass of crucible 3.94 g Mass of crucible + hydrated iron (II) sulphate 5.44 g Mass of crucible + anhydrous iron (II) sulphate (1st weighing) 4.72 g Mass of crucible + anhydrous iron (II) sulphate (2nd weighing) 4.65 g I will now use these results to do some calculations: Mass of hydrated iron (II) sulphate is 15.44 - 13.94 = 1.50 g Mass of anhydrous iron (II) sulphate is 14.65 - 13.94 = 0.71 g Mass of water present in the original sample of hydrated iron (II) sulphate is 15.44 - 14.65 = 0.79 g To find the moles of FeSO4 in the original sample you use: Moles = Mass / RFM Mass = 0.71 g RFM = 55.85 + 32.07 + (4 x 16) = 151.92 Therefore 0.71 / 151.92 = Moles = 0.004673512375 mol To find the moles of H2O in the original sample you use the same formula as above. Mass = 0.79 g RFM = (1 x 2) + 16 = 18 Therefore 0.79 / 18 = 0.043888888889 mol To find the value of (x) in the formula FeSO4.xH2O you need to find
Thermometric Titration Investigation
Thermometric Titration Investigation Sam Murphy 11T Planning We have to plan an experiment which measures the temperature change accompanying neutralisation so that it can be investigated. Neutralisation is when an acid and an alkali mixed together neutralise the other. The hydrogen and the hydroxide ions bond to form water and the rest bonds to form a salt. For example in the case of hydrochloric acid and sodium hydroxide, the salt is sodium chloride. Neutralisation is an exothermic reaction, like all reactions where bonds are formed. I plan to measure this heat energy released by the reaction (although it is possible to do the same for endothermic reactions, those that take heat energy from their surroundings rather than release). Neutralisation; Word Equation: Alkali + Acid = Salt + Water EXAMPLE: Sodium Hydroxide + Hydrochloric acid = Sodium Chloride + Water With titration equipment, I plan to add acid to an alkali (sodium hydroxide) until it is neutralised, Measuring the temperature so that I can work out the ? H. This means, I will add bit-by-bit, acid to the alkali in set amounts (3 ml) at a time. Measuring the temperature after adding the acid to the alkali. We will measure the change in Heat Energy by using this equation to work out the "? H" (Change in heat energy). For accuracy sake we will repeat the experiment for each acid so that we have two sets of
Identification of an Organic Unknown.
A-Level Chemistry Coursework Practical 1- Identification of an Organic Unknown Introduction In this investigation I will be supplied with a compound which will contain one of the following functional groups: * Alcohol * Aldehyde * Carboxylic acid * Ester * Ketone * Phenol The following is a flow diagram of the chemical tests I will use to identify the functional groups outlined above: (1.TEST WITH UNIVERSAL INDICATOR) TURNS RED TURNS ANY OTHER COLOUR (2.TEST WITH BROMINE WATER) (3.TEST WITH 2, 4 DNP) (4.TEST WITH TOLLENS REAGENT) General Safety Procedures For all experiments involving any unknown compound the wearing of goggles, lab coat and gloves is necessary because one or more of the compounds could be an irritant/corrosive/etc. . Test with Universal Indicator EQUIPMENT: Universal indicator Test tube Pipette PROCEDURE: Add several drops of the unknown substance to universal indicator inside a test tube using a pipette. OUTCOME: A colour change to red indicates either a carboxylic acid or phenol is present. Any other colour change shows alcohol, aldehyde, ketone or ester. EXPLANATION: Single indicators, such as litmus, are very weak acids. If the concentration of hydrogen ions is changed at a certain pH the indicator will
An investigation into the factors affecting the temperature rise of water heated electrically.
An Investigation Into the Factors Affecting The Temperature Rise of Water Heated Electrically Planning Aim: The aim of this investigation is to determine the factors affecting the temperature rise of water which is being heated electrically. Variables: The variables in this experiment are: ) Time 2) The Power Of The Heater 3) The Mass Of Water 4) The Temperature Rise The Possible Experiments There are several experiments which can be carried out to investigate the factors affecting water being heated electrically. ) Varying the mass of water used and the temperature rise and keeping the power of the heater and the time fixed. In this experiment the mass of water would be varied and keeping the power of the heater the same, the temperature rise could be record to produce an array of results. 2) Varying the power of the heater and the temperature rise, and keeping the mass of water and the time taken fixed. In this experiment the power of the heater would be varied and he again the temperature rise would be record to take results. 3) Finally the time could be varied with the temperature rise being varied, keeping the power and mass constant. In this experiment the time the heater was left on would be varied and the temperature rise would be recorded. From these 3 I have decided to choose 1) because number 3 is trivial, it is too simple and will not tell me
