The rate of reaction between sodium thiosulphate and hydrochloric acid Candidate name: Yeo Jin Kim (Kimberly Kim) Candidate number: 000791-021 Teacher: Yitao Duan 1.0 Aim The aim of this experiment is to investigate the changing of concentration effect the rate of reaction between sodium thiosulphate and hydrochloride. 2.1 Chemical Equation 2.0 Raw data 3.2 Qualitative Data 2.1.1 Before the experiment Substance | Color | State | Smell | Hydrochloride (HCl) | Colorless
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GCSE SCIENCE COURSEWORK (CHEMISTRY) PLANNING: Skill Area P Write down the aim of your Investigation. To investigate the effects of change in concentration on the rate of reaction between Sodium Thiosulphate and Hydrochloric Acid. Briefly describe your Preliminary Experiment and include your results. Explain how these results have helped you plan your main experiment (p8b). An increase in concentration will lead to an increase in the rate of reaction because there will be a greater number of
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evaporates‚ it lowers the temperature of the substances around them‚ because evaporation is an endothermic process. In the graph‚ each substance has a different curve. Some lowered the temperature much more than others. This is because different substances require differing amounts of energy to evaporate. To test this‚ the temperature change from the evaporation of 4 different alcohols was measured and compared‚ along with cyclohexane. Methanol had the largest change in temperature‚ 18.9 ºC. This indicates
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Abstract. This report is about how to standardize a Sodium Hydroxide (NaOH) solution by titrating it with pure sample of Potassium acid Phthalate (KHC8H4O4). This experiment has two sections. The first section is to standardize the Sodium Hydroxide by titration. Three sample of 0.7 – 0.9 g of solid KHP are place into each of the three numbered Erlenmeyer flasks. 50 ml of distilled water are added to each three of it from graduated cylinder and constantly shake it until the KHP solution are completely
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between hydrochloric acid and sodium thiosulphate is affected by the concentration of the acid Simple Procedure Place a conical flask on a piece of paper with a cross on it. Add hydrochloric acid and sodium thiosulphate‚ and record the amount of time taken for the cross to disappear through the solution from the top of the flask. Record this time and repeat this for different concentrations of hydrochloric acid. Fair Test The variables in this test are: The concentration of hydrochloric acid
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A buffer is an aqueous solution that allows resistance to significant change in pH when small amounts of strong acids or base are added to it. (Boundless‚ 2015) This is usually formed when a weak acid is added to a salt of its conjugate base. (Chemicool.com‚ 2014) When an Alka Seltzer tablet is dissolved in water‚ a buffer is formed when the weak acid citric acid is present with sodium citrate (citrate ion) the conjugate base (Buffer Balancing Acts Buffers‚ 2009) "Na3C3H5O7 (aq) + 3 HCl (aq)
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creating five reactions containing Sodium Sulfate that would confirm our compound and show its chemical properties. In each reaction‚ we replaced the presence of Sodium Sulfate with our unknown. Our first reaction was the reaction from the sulfate anion test between Sodium Sulfate and Barium Chloride. If the compound was in fact Sodium Sulfate it would produce a white precipitate and it did. The second reaction was first between Sodium Sulfate and Hydrochloric Acid‚ and then Silver Nitrate was added
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materials and soap making Soap making requires two major raw materials: * Fats and oils Fats and oils are esters of fatty acids and glycerol. These are obtained from natural substances which are classified into three: a. Nut oils Cold process soap making is the most suitable for this type of oil which is also characterized by large proportion of fatty acids. Nut oils saponify easily with strong alkali soulution. Example: coconut oil‚ palm kernel oil b. Hard fats Slow-lathering soaps
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Working out mean or average results 26.3+26.1+23.8= 76.2/3=25.4 The average ml of HCL needed in order to react for a colour change was 26.5ml. NaHCo3+ HCl NaCl+H2O+Co2 Calculation: 25 ml 0.1 mole dm3 26.3+26.1+23.8= 76.2 ml 76.2÷3=25.4÷1000=0.025 volume 0.1x0.25ml=0.025 moles 0.025÷0.254=0.98 (Rounded up to one decimals places) =0.1 M To begin with concentration I have choose three volumes form my results table which I have added them and then divided by three to get an average results. Describe
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Experiment 4: Analyze a solution of potassium hydroxide using standard hydrochloric acid Theory . Introduction: Titration is a procedure used in chemistry in order to determines the molarity of an acid or a base. In the other words‚ it is also consider as acid-base neutralization reaction (Darrell D. Ebbing 1976). A chemical reaction is set up between a know volume of a solution of unknown concentration and a known volume of a solution with a known concentration. The relative acidity or
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