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Cube compressive strength test
Cube compressive strength test Introduction: By far the most common test carried out on concrete is the compressive strength test. The reasons are that it is commonly assumed that most of the important properties of concrete are directly related to the compressive strength; since concrete has very little tensile strength, it is used primarily in a compressive mode, and therefore it is the compressive strength that is important in engineering practice; the structural design codes are based mainly on the compressive strength of concrete; and the test is easy and relative inexpensive to carry out. This time, we are going to use the three 150mm cube mold which we have made in last 2 week to test their compressive strength. Specimen: Three 150mm cubes (made on 14days ago) Apparatus: Compressive testing machine, calipers, weighing equipment Procedures: . The three cubes were removed from the curing tank. 2. Water and grit on the surface of the cubes were wiped off with a damp cloth. 3. Any faults and damages such as broken edge of the cube were visually inspected. 4. The dimensions of the three cubes were measured with calipers. No "under size" or "oversize" cubes were recorded. 5. The cubes were weighted 6. Test machine platen was wiped with a dry cloth. 7. The cube was placed centrally on the lower platen of the test machine with the rough as-cast top surface of the
Determine the concentration of a limewater solution
PLANNING AIM: To determine the concentration of a limewater solution. You have been provided with 250cm3 of limewater, which contains approximately 1g dm-3 of calcium hydroxide. Hydrochloric acid is also available, which has a concentration of exactly 2.00 mol dm-3. In order to determine the concentration of the limewater solution I will have to carry out a titration experiment between the acid being hydrochloric acid, HCl and the base being the limewater, Ca(OH) 2. However before continuing on with the titration you need to dilute the HCl acid since it is too concentrated to use. In order for you to do this you will need to work out the concentration of Ca(OH) 2. Balanced Equation of the titration: 2HCl + Ca(OH)2 --> CaCl2 + 2H2O 2 moles 1 mole 2 mol dm-3 ? Ca(OH)2 is 1g dm-3= 1 gram per 1dm3 No. of moles in Ca(OH)2 = Mass / Relative formula mass (RFM) = 1 / 74 = 0.0135 moles Concentration of Ca(OH)2 = No. of moles / Volume = 0.0135 / 1 = Ca(OH) 2 mol dm-3 Now that you have worked out the concentration of the lime water solution you can continue with the dilution process of HCl acid. In order to do this you will need to select a suitable concentration for the HCl acid to be diluted to. As we already know according to the balanced equation 1 mole of Ca(OH)2 reacts with exactly 2 moles of HCl,
Enthalpy change
Finding the enthalpy change that cannot be measured directly. Abstract The main purpose of the experiment was to plan and carry out experiments were the enthalpy change cannot be measured directly. And also to comment on the sources of error and assumptions made in the experiment. Introduction The standard enthalpy of formation of calcium carbonate, ?Hf is defined as the enthalpy change when one mole of calcium carbonate is formed from its elements in their standard states, under standard conditions. Equation for the reaction is: Ca (s) + C (graphite) + 1.5 O2 (g) --> CaCO3 (s) The reaction cannot be determined in the laboratory experimentally but by demining the enthalpy change when calcium carbonate reacts with hydrochloric acid, and applying the principles of Hess's law, this enthalpy change can then be measured. Safety Calcium is reactive with both water and acids. Care should be taken when water is added and eye protection worn at all times. Calcium is highly flammable. Apparatus * Eye protection * Polystyrene cup * 250 cm3 beaker * 100 cm3 measuring cylinder * -50 to 50 °C thermometer * 3 g calcium metal * 8 g of powdered calcium carbonate * 1.0 M HCL (aq) approximately 500 cm3 * Access to balance weighing ± 0.1g Method Experiment 1: Calculating the enthalpy change for the reaction between calcium metal and hydrochloric acid. . 1.03 g of
Finding Out how much Acid there is in a Solution
Finding out How Much Acid there is in a Solution When a metal is extracted from its ore, a waste product is often sulphur dioxide (SO2). This is then converted into sulphuric acid (H2SO4) and sold. To sell sulphuric acid, its accurate concentration must be known. In this investigation, I will use a titration method to attempt to find the accurate concentration of a sample of sulphuric acid thought to have a concentration of between 0.05 and 0.15 mol dm-3. The chemicals which I will use to do this will be solid anhydrous sodium carbonate (Na2Co3) and an indicator. Finding out How Much Acid there is in a Solution - Plan Quantities of Chemicals Required In this investigation, the chemical reaction used will be the neutralisation reaction between sulphuric acid and sodium carbonate: H2SO4 (aq) + Na2CO3 (aq) --> Na2SO4 (aq) + CO2 (g) + H2O (l) As the sodium carbonate has to be aqueous for this reaction, I will need to make up a solution of sodium carbonate before I begin the titration. I will need to choose an appropriate concentration for this solution and an appropriate volume of it to make up. As there is a 1:1 ratio between the moles of sulphuric acid and sodium sulphate used in this reaction, I should make up a solution of sodium carbonate of similar solution to that of the sulphuric acid. Assuming that the concentration of the sulphuric acid is around 0.1 mol dm-3, this
DETERMINING THE MASS OF LITHIUM
DETERMINING THE MASS OF LITHIUM METHOD 1 RESULTS Mass of lithium used 0.12g Amount of hydrogen gas produced 182cm³ TREATMENT OR RESULTS 2Li(s) + 2H2O(l) --> 2LiOH(aq) + H2(g) Moles of hydrogen gas collected Moles of H2 volume / 24000(cm³) 82 / 24000 0.007583333 moles Moles of H2 0.0076 moles Lithium that reacted H2 : Li : 2 0.0076 : 2 × 0.0076 0.0076 : 0.0152 Moles of Li 0.0152 moles Relative atomic mass of lithium R.A.M mass / moles 0.12 / 0.0152 7.894736842 R.A.M of Li 7.8947 METHOD 2 RESULTS Titration of aqueous LiOH with 0.100 mol dm-3 HCl. Start (cm³) End (cm³) Titre (cm³) 0 34.90 34.90 0 34.80 34.80 0 34.90 34.90 Average result (34.80 + 34.90) / 2 = 69.7 / 2 = 34.85 TREATMENT OF RESULTS LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l) Moles of HCl used in titration Moles of HCl concentration × (volume / 1000) 0.100 × (34.85 / 1000) 0.003485 moles Moles of HCl 0.0035 moles LiOH used in titration HCl : LiOH : 1 0.0035 : 0.0035 Moles of LiOH 0.0035 moles Number of moles of LiOH present in 100cm³ of solution from method 1 25cm³ of LiOH is pipette each time. There's 100cm³ of LiOH, so it will be: 0.003485 × 4 = 0.01394 Moles present in 100cm³ of LiOH 0.0139 moles Relative atomic mass of Lithium R.A.M mass / moles 0.12 / 0.0139 8.633093525 R.A.M of Li 8.633 HAZARD OF CHEMICALS IN THIS
Oxidation of ethanol
Name ________________________ Date ________________ Introduction For this exercise you are given full instructions for the practical procedure and these must be followed exactly. It is however up to you to consider suitable safety procedures and to organise your time appropriately. You should also carefully consider what is the most effective way of handling the materials and apparatus in order to obtain the maximum reliability. The Oxidation of Ethanol Ethanol is a primary alcohol and can be oxidised to either an aldehyde or a carboxylic acid. CH3CH2OH + [O] CH3CHO + H2O CH3CH2OH + 2[O] CH3COOH + H2O The purpose of this experiment is to oxidise ethanol and then to test the product to determine whether it has been oxidised to ethanal or oxidised to ethanoic acid. Read through the procedure below before starting the experiment. You should comment on the hazard of the chemicals use din the experiment. Method To 6cm3 of water in the pear-shaped flask, add 2cm3 of concentrated sulphuric acid, and set up the apparatus as shown below, but with a stopper in place of the dropping funnel. Ensure that all of the glass joints are greased. . Make up a solution containing 5g of sodium dichromate in 5cm3 of water, add 4cm3 of ethanol and pour the mixture into the dropping funnel. 2. Warm the acid in the pear shaped flask until it is almost boiling and turn off the
ion exchange
An experiment to purify contaminated water using ion exchange Introduction The aim of the experiment is to prove that there are positively charged copper ions (cations) contaminating a solution, which will be passed through a column of ion resin beads, with the product being pure water (or at least a less contaminated sample). Ion resin is made up of tiny beads, about 0.3-1.2mm in diameter. It's an insoluble substance, with a highly developed structure of pores - which can either trap of release ions. This is known as ion exchange, a process of exchanging ions from two substances, an insoluble solid and a solution. An example of this process is softening hard water and water purification. When a solid is comes into contact with a solution containing ions, equilibrium is formed - meaning that the reaction can go either way, it's a reversible reaction. Ion exchange occurs when the reaction of two compounds (or elements) exchange their ions to form a new structure in a solution. An ion is a charged atom/molecule, which has either lost or gained an electron, therefore giving it a positive or negative charge. (http://en.wikipedia.org/wiki/Ion) Hypothesis It is expected, that it will be possible to purify contaminated water using ion exchange Safety assessment Goggles Prevents from any splashes getting in the eyes. Lab coats Prevents staining of clothes. Stools and
Synthesis of Aspirin
The aim of this practical is to carry out a chemical reaction between salicylic acid with anhydride. This will allow us to calculate the percentage yield. The chemical company for which you work has developed a new laboratory method for synthesising aspirin which produces a high yield of pure product. Before investing the large amount of money it would take to scale up the process for mass production of aspirin, the directors want to know if this new method is significantly better than existing methods, in terms of the percentage yield and purity of aspirin produced. Theory Aspirin, which is sometimes called acetylsalicylic acid, was first made in 1893 by a German chemist Felix Hofmann. Aspirin is classified as organic ester and organic acid. It is mostly used in medicine like pain killers and also they are used for reducing fever. Aspirin is made by reacting salicylic acid with acetic anhydride. The human generation already knew how useful salicylic acid for pain. It was until the 19th century when we learned how to make mass production of aspirin and how to cope with the toxic side effects. Aspirin has many pharmaceutical uses; the most important is the cure for any kind of pain. It can also treat arthritis because it has an anti-inflammatory effect and also reduce fever. One of the majors side effects gastritis this gives stomach aches. Mass/g Weighing boat + 3
Which Equation is Correct
Which Equation is Correct? In this investigation, I am going to plan an experiment to determine the correct equation for the decomposition of basic copper carbonate. Basic copper carbonate is a mixture of 60% CuCO3 and 40% Cu(OH)21. It has two oxides, Cu2O and CuO. When it is heated, basic copper carbonate decomposes to form one of its oxides. The gases produced during the decomposition come from the CuCO3. The equation for possible reactions can be either of the following: Equation 1: 2CuCO3(s) Cu2O(s) + 2CO2(g) + 1/2O2(g) Equation 2: CuCO3(s) CuO(s) + CO2(g) To determine which of these 2 equations are correct, I will be decomposing 0.4g (to nearest 0.01g) of basic copper carbonate. My predictions for both of the equations are below: Equation 1: Ratio of - 2CuCO3 : 2CO2 + 1/2 O2 1 : 1.25 Mass of 1 mole of CuCO3 = 123.52 g Mass of Copper Carbonate used = 0.4g Moles of 2CuCO3 = 0.4 ÷ 123.5 = 0.003 moles As the ratio is 1: 1.25, so Moles of Gas given off = 0.003 × 1.25 = 0.0037 moles At room temperature (20°c) & pressure (1atm), 1 mole of gas = 24000cm3 So if the equation 1 is correct, then volume of gas given off = 0.0037 × 24000 = 89cm3 Equation 2: Ratio of - CuCO3 : CO2 1 : 1 Mass of 1 mole of CuCO3 = 123.5g Mass of Copper Carbonate used = 0.4g Moles of CuCO3 = 0.4 ÷ 123.5 = 0.003 moles As the ratio is 1:1, so moles of gas given off = 0.003