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International Baccalaureate: Chemistry

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  1. Paper Chromatography on Amino Acids IA

    0.74 0.73 Leucine Looking at the data above, although the Rf values obtained differs slightly from the literature values, we could see that both values are only 0 to 0.02 apart. Both sets of values (obtained and literature) show that Leucine has the highest Rf value and lysine has the lowest value with aspartic acid in between. The actual amino acids were meant to be lysine aspartic acid and leucine. This means that my experiment was successful as I was able to identify all three amino acids correctly.

    • Word count: 1412
  2. Which fruit juice (Orange, Apple, Peach, Pineapple, and Apricot) has the highest concentration of ascorbic acid?

    Graduated cylinder 3. 100 milliliters of dichlorophenolindophenol (DCPIP) indicator solution 4. Burette 5. Distilled water 6. Funnel 7. Conical flask 8. Beakers PROCEDURE: 1. Gather your materials. 2. Set Burette onto stand with clamp. 3. Fill a graduated cylinder with 10 mL of distilled water and pour into a beaker. 4. Fill a graduated cylinder with 10 mL of Apple Juice and pour into the beaker with distilled water. 5. Mix the solution and put the solution in a 100 mL conical flask. 6. Put the conical flask aside.

    • Word count: 1091
  3. Reactivity of Metals with Water and Acid

    1. dilute hydrochloric acid (0.5 mol/L) 2. spark lighter 3. Bunsen burner clamped to a retort stand 4. paper towel 5. wooden splint 6. masking tape or test-tube stopper Procedure Part A: Reactivity of Metals in Water The materials were selected and brought to the station. Safety protocol was followed and safety glasses were worn throughout the experiment. The appropriate data tables were drawn in which to record the data observations. The beaker was filled with approximately 250 mL of tap water. The water was tested with pH paper and the results were recorded. On a folded piece of paper towel, the professor placed a small piece of calcium (Ca).

    • Word count: 1929
  4. In this lab, we will use the sodium hydroxide to determine the percent acetic acid in vinegar using titration.

    In titration labs we assume that the number of moles of the base is the same as the acid?s: 1. 2. 3. Now we multiply the volume by the molarity: 1. 2. Got the molarity from the previous lab 3. Assuming that the moles of OH- = H+ we just multiply the moles by the molar mass to give the mass of acetic acid: 1. 2. 3. Now we divide the mass of acetic acid by the mass of the vinegar and then multiply it by a hundred to get the percentage.

    • Word count: 1254
  5. Effect of temperature on pH of water samples

    Small changes in pH can endanger many kinds of plants and animals; for example, trout and various kinds of nympH s can only survive in waters between pH 7 and pH 9. If the pH of the waters in which they live is outside of that range, they may not survive or reproduce. Changes in pH can also be caused by algal blooms (more basic), industrial processes resulting in a release of bases or acids (raising or lowering pH ), or the oxidation of sulfide-containing sediments (more acidic).

    • Word count: 1670
  6. Calculate % of caco3 in white egg shell

    Dry your egg shell with a paper towel and put it into a beaker. 3. Dry the shell further either in an oven or by using hot air from a hair dryer. 4. Grind the shell to a fine powder with a mortar and pestle. For the next few steps your setup should look like this: 1. Weigh accurately 0.5g of the power in a conical flask. 1. Use a mortar and pestle to powder the eggshell. Powdering will increase the rate of reaction and will result in the time for the experiment to occur to reduce.

    • Word count: 1166
  7. Percent Yield Lab. This experiment has proven that KI is the limiting reagent in this chemical reaction.

    This calculation is performed twice once for the first reactant and one for the second. Then the Theoretical Yield is calculated, it is how much product will be synthesized with the reactants. Multiply the lowest number of moles (limiting reagent?s mole) by the molar mass of the product. This will give you the mass of the product. In order to find the percent yield, a ratio between the actual yield and the theoretical yield is used. This indicates the percent of how much of the theoretical yield was obtained in the experiment.

    • Word count: 1671
  8. Experimental Molar Enthalpy of Neutralization for Sodium Hydroxide Solution

    Using this information, a calorimetric lab was conducted to find the molar enthalpy of neutralization for the sodium hydroxide solution. Through molar enthalpy calculations, the experimental molar enthalpy of neutralization for the sodium hydroxide solution was found to be -64.0±3.3KJ/mol; however, the theoretical (actual) molar enthalpy of neutralization for the sodium hydroxide solution is -57KJ/mol. In other words the experimental enthalpy change was -64.0±3.3KJ and the theoretical (actual) enthalpy change was -57KJ. This as a result produced a 12% difference.

    • Word count: 1351
  9. Finding thr Percentage Composition of Magnesium Oxide

    Experimental Design You will burn two identical strips of magnesium in ceramic crucibles. The contents of one crucible will be analyzed as they are. In the other crucible, any magnesium nitride that may have formed will be chemically converted into magnesium oxide. You will then determine the mass of the product in both crucibles and use this value to calculate the percentage composition of magnesium oxide. You will compare the two calculated values. Materials 1. 2. Chemical safety goggles 3. Ceramic crucible and lid 4. Balance 5. Ring clamp 6. Bunsen burner clamped to the retort stand 7.

    • Word count: 1496
  10. Determining Ka by the half-titration of a weak acid

    Thus, we get: = PKa + = PKa Now the PH was, so PKa= 5.0 ±0.1 = 5.0 ± 2% 5.0 ±2% = = 10-5 ±2% Titration curve: To get error we are going to sketch a titration curve, and from this measure the PH at half equivalence. To do this: PH of acetic acid (1M): Ka = = 10-4.76 = √(1×10-4.76) So PH of acetic acid= 2.38 Now PH of NaOH, (1M) Now concentration of NaOH, was 1M So = 1 = -log(1)

    • Word count: 1985
  11. Chemistry: Strong Acid and Weak Base Titration Lab

    Therefore, the following relationship holds: nVb x Cb = Va x Ca Where: Vb = the volume of the base Cb = the concentration of the base Va = the volume of the acid Ca = the concentration of the acid n = the mole factor In the case of hydrochloric acid and Sodium Bicarbonate (Baking Soda), the mole ratio is one to one, thus the mole factor is 1. Therefore, the volume of sodium bicarbonate multiplied by its concentration in molarity is equal to the moles sodium bicarbonate.

    • Word count: 1803
  12. Chemistry Investigation to find the Empirical Formula of Magnesium Oxide

    Total weight before ± 0.001 (g) Total weight after ± 0.001 (g) 1 34.333 34.452 33.449 2 37.841 38.005 38.044 4 33.834 33.933 33.234 5 35.262 35.397 35.453 6 33.596 33.719 33.789 Trial 3 Group* Weight of crucible ± 0.001 (g) Total weight before ± 0.001 (g) Total weight after ± 0.001 (g) 6 31.687 31.773 31.806 *Missing numbers are due to incomplete number of trials by the groups and/or the results lacked in validity Data Processing From the data collected from the reaction of magnesium (mg)

    • Word count: 1418
  13. Research into the production of Nitrate Fertillisers.

    Disadvantages of ammonium Nitrate Fertilizer As I said earlier, ammonium nitrate is a strong and explosive agent. In fact it is one of the largest industrial explosive manufactured in the US. Ammonium nitrate is used in the fields of quarrying and mining. Ammonium nitrate is one of the cheapest crop nourishing fertilizer types and hence, it is easily available in the markets and agricultural stores too. Chemical formula for ammonium nitrate is (NH4)2SO4 Chemical Properties ammonium nitrate Chemical formula: NH4NO3 Composition: 33 to 34% N Water solubility (20 ºC): 1,900 g/L Granular ammonium nitrate provides equal amounts of nitrate-N and ammonium-N, and its application has been highly suited to vegetable or forage crops.

    • Word count: 1001
  14. Experiment Plan. Chemistry IA: Electrolysis of Metal Sulphate solutions (NiSO4)

    Therefore indicating that the nickel sulphate solution is ionised by the electric current and dissociated into nickel ions and sulphate ions. This can be represented in a chemical equation: NiSO4 ï Ni2+ + SO42- At the cathode, positively charged nickel ions are formed there and Ni2+ ions are reduced to Ni by gaining two electrons: Ni2+ + 2e ï Ni At the anode, Ni is oxidised into Ni2+ by dissolving and going into the nickel sulphate solution and finally depositing nickel at the cathode: Ni ï Ni2+ + 2e When the electrolysis circuit has electricity flowing, the nickel ions will float towards the electrode.

    • Word count: 1261
  15. Lab report. Finding the molar enthalpy change of the reaction between Hydrochloric acid and sodium carbonate

    I then found the temperature on this line when the temperature started increasing so finding the value of were the peak would have been if it wasn?t for heat loss. This is one of my graphs: Now having the results I proceeded to find the enthalpy change for the reaction I did this carrying out number of steps, which were: I started off by calculating the average temperatures for each experiment I did this by using the formula: . I determined the uncertainty as according to textbook the uncertainties when calculating averages is the biggest difference between the average and one of the values used, I so got: Experiment Average temperature change (k)

    • Word count: 1549
  16. Hypo Sodium Thiosulfate Kinetics Lab

    Our reaction: Can be manipulated, rearranged, and balanced to find the net ionic equation of the reaction: This reveals to us that sodium and chlorine are spectator ions, and the majority of the change of the reaction is done to the sulfur-related ions. This, however, does not alone reveal our rate or rate law. We will determine later on why this isn?t possible. Procedure: Obtain a reservoir of75mL of .17 M sodium thiosulfate. This amount will be enough for all five of your trials.

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  17. Neutralization Titration using a Strong Acid and Weak Base

    the volume of Sodium Carbonate in each titration 2. the mole of the Hydrochloric acid used 3. the mole of the Sodium carbonate solution used 4. the indicator used (Methyl Orange) Materials and Equipment Part A 1. Anhydrous Sodium Carbonate (NaCO3) 2. Deionized water 3. 100 cm3 Beaker 4. 250 cm3 Volumetric Flask with stopper 5. Small Funnel Part B 1. Volumetric flask of 250 cm3 NaCO3 from part A 2. 20.00 cm3 pipette 3. Methyl orange indicator 4. 50 cm3 Burette 5. Small Beaker 6. Hydrochloric Acid (HCl) 7. 100 cm3 Conical flask(s) Diagram Method Part A 1.

    • Word count: 1245
  18. Lab Report Determining The Relative Molecular Mass of Amidosulphuric Acid

    Amidosulphuric acid in pipette 0.03 x 100% = 0.12 % 25.00 Sodium hydroxide used 0.10 x 100% = 0.4% 25.03 Average Volume of Sodium Hydroxide used =25.10cm³+25.10cm³+25.20cm³ 3 = 25.03cm³ Standardization equation H2NSO3H + NaOH H2NSO3Na + H2O From the equation we can see that 1 mole of amidosulphuric acid (H2NSO3H) reated with 1 mole of sodium hydroxide (NaOH) to form one mole of H2NSO3Na and one mole of water (H2O). Number of mole, n for sodium hydroxide n = MV 1000 = 0.096M x 25.03cm³ 1000 = 0.0024 mole ± 0.4% From the equation,1 mole of sodium hydroxide reacted

    • Word count: 1142
  19. How the concentration of hydrochloric acid affects the rate of the reaction with magnesium.

    An increased number of reactants mean there will be a higher probability of successful collisions (collisions with energy greater than activation energy and correct geometry of collisions) resulting in more reactants being converted into products. So as concentration is increased a greater volume of gas will be evolved in a shorter amount of time than when the concentration is lower. Materials 1. 100 ml gas syringe 2. 25 ml of hydrochloric acid; 0.5M, 1.0M, 1.5M, 2.0M, 2.5M 3. Side armed flask 4.

    • Word count: 1663
  20. Calculating Density -How can you find the volume and density of two regular solids, two irregular solids, and two liquids?

    2. 2 Irregular Solids (Cannot be the same) 3. 2 Liquids (Cannot be the same) 4. 1 Triple Balance Scale 5. 25 mL Graduated Cylinder 6. Distilled Water 7. Safety Goggles 8. Apron 9. Close Toed Shoes 10. Ruler (foot long) Procedure: 1. Measure the length of the object in centimeters. 2. Measure the width of the object in centimeters. 3. Measure the height of the object in centimeters. 4. Multiply length by width, and then multiply the product by the height. 5. Fill the beaker up to a testable amount in which the object may be placed in without the distilled water overflowing.

    • Word count: 1265

    They are all rather soluble in dilute acids. The compound to be prepared is potassium tris(oxalato)ferrate(III) trihydrate. It is an octahedral transition metal complex in which three bidentate oxalate ions are bound to an iron centre. The general equation of the reaction is; (NH4)2[Fe(H2O)2(SO4)2]*4H2O + H2C2O4*2H2O FeC2O4 + H2SO4 + (NH4)2SO4 +8H2O H2C2O4*H2O + 2FeC2O4 + 3K2C2O4*H2O + H2O2 2K3[Fe (C2O4)3*3H2O + H2O A titration with potassium permanganate (KMnO4) will then be used to determine the amount of metal in the oxalate complex.

    • Word count: 1537
  22. Discovering the formula of MgO

    Magnesium Oxide, depends on how much Magnesium is added. The dependent variable will be the mass of Magnesium Oxide, this is a measured variable as the mass of oxygen can be calculated and will enable us to determine the formula for Magnesium Oxide. Controlled variables: 1. Container used 2. Surface area of Magnesium 3. Concentration of oxygen in the container 4. Temperature of flame Controlling variables: 1. A crucible with a lid is used and is filled with a layer of filter paper, in order to allow combustion in a closed environment, preventing the loss of Magnesium oxide powder. 2. Magnesium ?ribbon? will be used in all cases. 3.

    • Word count: 1614
  23. We can conclude that the experiment gave a reasonably accurate value for the number of moles of water of crystallization in oxalic acid

    This experiment was repeated three times with the same variables and measurements to get an average of the readings collected for more accurate results. Quantitative Data: RAW DATA Experiment Mass of weighing bottle (+/- 0.01g) Mass of weighing bottle and oxalic acid crystals (+/-0.01g) Mass of oxalic acid crystals (+/- 0.02g) Volume of oxalic acid solution (+/- 0.3cm³) Start volume of burette (+/- 0.5cm³) End volume of burette (+/- 0.5cm³) 1 13.4 14.9 1.5 25.0 0.0 24.6 2 13.4 14.9 1.5 25.0 0.0 24.8 3 13.4 14.9 1.5 25.0 0.0 24.7 PROCESSED DATA Experiment Volume of oxalic acid solution (+/- 0.3cm³)

    • Word count: 1132
  24. Chemistry Lab Report- Determining the Enthalpy of Enthalpy Change, H, for a Redox Reaction (

    Trial 1 113.3g Temperature Change of CuSO4 Solution (°C)(±0.5°C) Trial 1 0°C Temperature Change of Cu + Zn2+ solution (°C) (±0.5°C) Trial 1 38°C Mass of CuSO4 Solution (g) (±0.05g) Trial 2 25.0 grams Mass of Zinc Powder (g) (±0.05g) Trial 2 4.0 grams Mass of Copper Calorimeter (g) (±0.05g) Trial 2 113.3g Temperature Change of CuSO4 Solution (°C)(±0.5°C) Trial 2 0°C Temperature Change of Cu + Zn2+ solution (°C) (±0.5°C) Trial 2 38°C Average of Temperature rise (ΔT): 38°C+38°C 2 trials= 38°C This value will be used to calculate the enthalpy change throughout this experiment.

    • Word count: 1153
  25. Research report on Stoichiometry

    It is the part of chemistry that studies amounts of substances that are involved in reactions. Stoichiometry is the branch of chemistry and chemical engineering that deals with the quantities of substances that enter into, and are produced by, chemical reactions. The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). Stoichiometry deals with calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction. WHY, WHAT AND HOW All reactions are dependent on how much stuff you have.

    • Word count: 1040

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