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International Baccalaureate: Chemistry
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We took 10ml of HCl in the conical flask and added 2-drops of phenolphthalein indicator and then we titrated it by taking standardized solution of NaOH in the burette. The end point of the titration was again the same the colour change from colourless to (pink), we recorded the volume of NaOH consumed and then did the calculation to figure out the exact molarity of the HCl. Part C: (Amount of Calcium Carbonate present in the Egg-Shell) The experiment was done in three steps:- By removing the egg shell from the boiled egg, we collected the shell in a watch
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Lab report - Determine the Empirical Formula of Magnesium Oxide by reacting a known mass of Magnesium with Oxygen.
Data collection and Processing Quantitative data: Apparatus Mass (±0.1g)* Crucible and lid. 39.8 Magnesium ribbon. 1.5 Crucible, lid and magnesium. 41.5 *(Uncertainty determined from the last decimal place of the electronic balance) Processed Data Apparatus Mass (g) Magnesium oxide (MgO) 1.7(+/-0.2) Oxygen (O) 0.3(+/- 0.3) Calculations: MgO = 41.5(+/- 0.1g) - 39.8(+/-0.1g) = 1.7(+/-0.2g) Oxygen = 1.7(+/-0.2g) - 1.5(+/-0.1g)= 0.3(+/- 0.3g) Deducing the Empirical Formula. Process Magnesium Oxygen n= m/M 1.5(±0.1g)/24.3g/mol = 1.5(±7% g)/ 24.3g/mol = 0.0617(±7%) 0.3(±0.3g)/16.0g/mol = 0.3(±100%)/16.0g/mol = 0.1875 (±100%) Simplest ratio 0.0617(±7%)/0.01875(±100%) = 3.3(±107%)
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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)
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Acidic solutions make the hair seem smoother. Basic solutions make their hair seem frizzier. Test B (shake test- determination of foam formation): Put approximately 10ml of the 1% shampoo solution into a 250ml graduated cylinder and record the initial volume of the solution. Cover the cylinder with your thumb and shake 10 times. Record the total volume of the contents after shaking. Then calculate the volume of the foam only by subtracting the initial volume of the solution without the foam. Information: The smaller the bubbles the better the shampoo. Test C (foam quality and retention): This test should be done together with the previous test.
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The addition of the Potassium Permanganate crystals to the water and agar, there will be a result of more diffusion and more obvious color change in water compared to agar which will result in a slower less obvious diffusion. Materials The materials needed for this experiment are: One Petri dish with 2% agar and one Petri dish of water filled half way up to test the rate of diffusion. Also, two small crystals of potassium permanganate and one pair of forceps will be needed to place in the Petri dishes.
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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.
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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.
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EXAMPLE: 56 Fe 26 ? The mass number = protons + neutrons = 56 The atomic number = number of protons = 26 Therefore the number of neutrons = 56-26 = 30 Isotopes are atoms of the same element with different mass numbers (atoms of the same element must contain the same number of protons, but can contain a different number of neutrons) Be able to identify protons, neutrons and electrons in isotopes and in ions. For a given element the number of protons never changes The number of neutrons depends on the mass number of the isotope in question The number of electrons is the same as the number of protons UNLESS we are dealing with an ion!
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Periodic Table The number of electrons in the highest occupied energy level are the same number as the period the element is in. Define the terms first ionization energy and electronegativity. First ionization energy is the amount of energy needed to remove one mole of electrons from one mole of gaseous atoms. Electronegativity is the attraction of a covalently bonded atom for a bonding pair of electrons. Describe and explain the trends in atomic radii, ionic radii, first ionization energies, electronegativities and melting points for the alkali metals (Li ï Cs)
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As time elapsed, the stream has diminished. The only mistake was that a small piece of magnesium was dropped into the beaker containing water. A stream of small bubbles was emitted from the piece of magnesium sunk to the bottom. No bubbles flew out of the burette and no obvious mistakes were made. Calculation The number of moles of hydrogen gas produced is calculated through the reaction that it undergoes: The molar volume of hydrogen gas at STP is calculated using the combined gas law: According to the combined gas law: The molar volume of hydrogen gas at STP is Theoretical value: Molar volume of hydrogen gas Percentage error 100% 4.02% Trial 2 Raw Data Table Mass of magnesium ()
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Faint green tinge became evident and fine white powder was formed from crystals. Continued heating led to formation of oliver green colour. Quantitative data – Original weight of hydrous copper sulfate – 3g Error in electronic balance - ± 0.001g (provided by the teacher) Data collection – Trial 1 1.96g Trial 2 1.87g Trial 3 1.94g Data Processing – To minimize the errors I conducted three trials and averaged the readings to find a more accurate answer. = 1.93 g ± 0.001 Change in mass = (3 ± 0.001) – (1.93 ± 0.001)
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= 25cm3 Uncertainty in volume (âV) = ± 1 cm3 Mass of CuSO4 taken (M) = 25g Uncertainty in weighing scale (âM) = ± 0.01 g Mass of Zinc taken (m) = 2g Uncertainty in mass of zinc (âm) = ± 0.01 g Specific Heat capacity of CuSO4 (c) = 4.184 J/g°C Data Processing Graph of temperature Vs time when zinc powder was added to Copper Sulfate Intitial temperature (t1) = 27 °C Final temperature (t2) = 65 °C Change in temperature (âT)
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Investigating Factors that Affect the Rate of Reaction of the Decomposition of Hydrogen Peroxide (Only method and tables no calculations0
Dependent Variable- The rate of reaction of the decomposition of hydrogen peroxide ? rate of reaction = . Research Question- it is needed to calculate the rate of reaction (kPa sec-1) of the decomposition of H2O2 to understand how different factors such as the change in concentration and the change in temperature of H2O2 affect the rate of reaction. Materials and Method- Materials: 0.5 M Yeast solution (the catalyst) - 15 mL 45 mL of 3 % H2O2 solution A thermometer A computer with LoggerPro Program. A Vernier computer interface A Vernier Gas Pressure Sensor A 1 liter beaker A match to light up the bunsen burner A tripod Two 10 mL test tubes Two 10 mL pipette
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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)
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radiation from the sun breaks the bonds of molecules to form atoms and free radicals. Temperature in the ionosphere is about -50ºC. Point 4.2 – Identify the main pollutants found in the lower atmosphere and their sources. Here is a table showing the main pollutants found in lower atmosphere (troposphere) and their sources: Pollutant Sources – artificial (caused by humans) Sources – natural Nitrogen oxides (NOx) e.g. nitrogen dioxide, nitric oxide The high temperature environment in internal combustion in vehicles. N2 (g) + O2 (g) ï 2NO (g) Burning of biomass, heat caused by lightning, soil bacteria. Carbon monoxide Incomplete combustion in motor vehicles C8H18 (l) + 5O2 (g) ï CO( g) + 7C (s)
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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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Determining the Percentage Yield of Sodium Chloride produced from Sodium Bicarbonate and Hydrochloric Acid
Material: 1. Sodium bicarbonate 2. Test tube rack 3. Bunsen burner 4. Electronic balance 5. Test tubes x 3 6. Sodium Chloride 7. Test tube clamp Method: At the beginning, a table has been made in order to record all the data. After the test tubes were numbered from one to three, they were weighted to 0.001 grams. Later, sodium bicarbonate was added to each test tube and they were again separately weighted.
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Research question: how to convert NaOH to NaCl by two different routes , and measure the enthalpy changes for each one in order to test Hesss law ?
ΔT = the change in temperature (ºC) Given the volume and concentration of each solution, the number of moles for each substance in the reaction can be calculated, therefore, the enthalpy of each reaction can be calculated in materials needed : 1. ( 8.0 grams )sodium hydroxide, NaOH 2. labQuest 3. 2 graduated cylinders of size( 100 mL ± 0.5) 4. (100 cm3) 2 M Hydrochloric acid solution ,HCL 5. Measuring pipette (25cm3) 6. glass stir stick 7. mortar 8.
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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.
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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
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At the endpoint: 2(no. mole of acid) = no. of mole of base 2(Molarity of acid*Volume of acid) = Molarity of base*volume of base Trail 1: Molarity of acid *10/1000 = (0.5* 12.5/1000)/2 Molarity of acid =0.3125M Trail 2: Molarity of acid *10/1000 = (0.5* 12.2/1000)/2 Molarity of acid = 0.305M Trail 3: Molarity of acid *10/1000 = (0.5* 12.45/1000)/2 Molarity of acid = 0.31125M Trail 1 Trail 2 Trail 3 Final burette reading(ml) 12.50 24.70 37.15 Initial burette reading(ml)
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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.
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The aim of this experiment is to examine the enthalpy of combustion of the first five alcohols and to see how does it change with the increasing chain length
Hypothesis : The longer the chain , the higher the heat of combustion . Explanation of the expectations : The more bonds there are holding the carbon, oxygen and hydrogen atoms together, more energy will be required to break them apart. For example Ethanol has the formula C2H5 OH. In this formula there are five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To separate these types of bonds certain amount of energy is required , which will be shown later.
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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.
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Volume in the volumetric flask : 250 + 0.5 cm³ 3. Volume of the pipette solution : 25 + 0.5cm³ Data collection : Quantitative data : Table 1.0 : Results of oxalic acid used in the experiment. 1 2 3 Initial burette reading (cm³) + 0.05cm³ 0.00 0.00 0.00 Final burette reading (cm³) + 0.05cm³ 29.60 29.00 27.50 Volume of H2C2O4.2H2O used (cm³)
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