552 results found

#### Calculating the specific heat of a metal

The Specific Heat of a Metal I. Purpose: to determine the specific heat of a substance. II. Materials: > 50-mL beaker > 250-mL beaker > 400-mL beaker > 100-mL graduated cylinder > Large test tube > Glass stirring rod > Utility clamp > Ring stand > Ring support > Hot plate > Electric balance > Plastic foam cup > Thermometer > Lead shot > Distilled water III. Procedure: . 250 mL of water was heated in a 400-mL beaker until it was boiling gently. 2. While the water was heating, the mass of a clean, dry 50-mL beaker was determined and recorded. Between 80 g and 120 g of lead shot was then added to the beaker and their combined mass was measured and recorded. 3. The lead shot was then transferred to a large, dry test tube. The utility clamp was used to suspend the test tube in the boiling water; the lead shot was below the level of the water in the beaker. The test tube was then left in the boiling for 10 minutes. 4. While the lead shot was heating, 100 mL of distilled water was measured I na graduated cylinder. The water was poured into a plastic foam cup that was placed in a 250-mL beaker for support. 5. The temperature of the water in the plastic cup and the water in the boiling bath was measured. 6. The test tube was removed from the boiling water and quickly poured into the water-filled plastic foam cup. A thermometer and a glass stirring rod were

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• Word count: 583
• Level: International Baccalaureate
• Subject: Chemistry

#### Factors affecting Galvanic Cells

Factors affecting Galvanic Cells Problem: If and how changes in the solutions as well as their concentrations affect the voltage in a galvanic cell? Variables: - Independent: * Type of solution used: Cu(NO3)2, Zn(NO3)2, Pb(NO3)2 * The concentration of the each solution. - Dependant: * The mass change in the cathode and anode - Controlled * The amount of time * The metals used for both rods * The temperature of the solutions Materials: - Power supply - Voltmeter - Zinc and Copper metal rods - Circuit wire - Beakers - Salt Bridge - Sand Paper - Crocodile clips - Graduating Cylinder - Cu(NO3)2, Zn(NO3)2, Pb(NO3)2 - Weight Scale Procedure: - Set up a beaker with 300 mL of 1M Cu(NO3)2 - Set up the voltmeter so it's connected to the power supply, correctly and with crocodile clips - Set up the beakers and the salt bridge as shown in the diagram - Obtain small strips of filter paper, to be used as salt bridges. Wet each strip with Cu(NO3)2 and insert into the solution. Repeat for each of the three remaining salt solutions. - Start the power supply for 5 minutes and record the amount of voltage created. Do so three times and record the information - Repeat above steps for 0.5M, 1.5 M and 2M and then repeat so with the remaining

• Word count: 238
• Level: International Baccalaureate
• Subject: Chemistry

#### THE CATALYTIC DECOMPOSITION OF HYDROGEN PEROXIDE

Chemistry Internal Assessment: THE CATALYTIC DECOMPOSITION OF HYDROGEN PEROXIDE Introduction: For this internal assessment, I will be looking into the catalytic decomposition of hydrogen peroxide. I will be conducting an experiment to show how the mass of catalyst used affects the rate at which hydrogen peroxide decomposes. My research question will be, "How does the mass of Manganese (V) Oxide used affect the rate of decomposition of hydrogen peroxide. A catalyst is a substance which alters the rate of a chemical reaction but is chemically unchanged at the end of the reaction. This means that the amount of catalyst at the start and the end of the reaction will remain constant. Catalysts alter the rate of a chemical reaction and this is the reason why I have chosen to conduct an experiment to find out to what extent a catalyst could affect the rate of a chemical reaction. Aim: To investigate how the mass of catalyst (manganese (V) oxide) used affects the speed of decomposition of hydrogen peroxide, H2O2. Hypothesis: My hypothesis for this experiment is that the more catalyst I used, the faster the decomposition of the hydrogen peroxide. This is because catalysts help to overcome activation energy and the more catalyst used, the more activation energy will be overcome and thus the reaction will be faster. However, I also expected that eventually there would be a point

• Word count: 2137
• Level: International Baccalaureate
• Subject: Chemistry

#### determining the empirical of magnesium oxide

MAGNESIUM OXIDE-EMPIRICAL FORMULA CHEMISTRY HL JAIME CASTRO A. 0-2 PRESENTED TO: KEITH RIGBY ANGLO COLOMBIAN SCHOOL EXPERIMENTAL SCIENCES DEPARTMENT RESULTS: The following table shows the mass recorded for the crucible, the magnesium and the magnesium oxide. Table1. Showing the masses recorded during the practical Crucible /g ±0.01g Crucible and magnesium /g ±0.01 g Crucible and magnesium oxide /g ±0.01g 25.50 25.67 25.80 If we subtract the mass of the crucible from the masses recorded in table 1 we can get the values of the masses of the magnesium and the magnesium oxide, the next table shows the exact masses of both magnesium and magnesium oxide. Table2. Showing the mass of Magnesium and the mass of magnesium oxide Magnesium /g ±0.01 g Magnesium oxide /g ±0.01g 0.17 0.30 ANALYSIS OF RESULTS: Taking in count the data recorded in table 2 we can calculate the mass of the oxygen from the air that reacted with the magnesium in order to form magnesium oxide, this can be done by just finding the difference between the values of the two masses, like shown below. Mass of Magnesium: 0.17g(±0.01 g) Mass of Magnesium oxide: 0.30g(±0.01 g) Mass of oxygen = (Mass of Magnesium oxide - Mass of Magnesium) = (0.30 - 0.17) = 0.13g Mass of oxygen = 0.13g Now that we know that the mass of the oxygen involved in the reaction we

• Word count: 808
• Level: International Baccalaureate
• Subject: Chemistry

#### Hess's Law. The experiment conducted was meant to determine the enthalpy of formation of MgO(s) and CaO(s)

Hess's Law Data Collection and Processing Table V: Heats of Formation Heat of Formation (kJ/mol) Substance Trial 1 Trial 2 Magnesium oxide - 101.5 ± 0.5 - 81.1 ± 0.5 Magnesium - 366.1 ± 0.4 - 322.2 ± 0.4 Calcium oxide N/A - 141.7 ± 1.0 Calcium - 192.6 ± 0.5 - 173.6 ± 0.5 Notes: * The heats of formation were calculated using the formula ?H = Q/nlimiting reactant Ex: Magnesium oxide Trial 1: n(MgO) = (1.08 g ± 0.01 g)/40.304 g/mol = 0.026796 mol n(HCl) = (100.0 g ± 0.5 g)/36.461 g/mol = 2.7427 mol MgO 1 mol 0.026796 mol ¯¯¯¯= ¯¯¯¯¯ = ¯¯¯¯¯¯¯¯¯¯¯¯ H2O 1 mol 0.026796 mol HCl 2 mol 2.7427 mol ¯¯¯ = ¯¯¯¯¯ = ¯¯¯¯¯¯¯¯¯¯ H2O 1 mol 1.3714 mol Therefore MgO is the limiting reactant. Assuming HCl(aq) has the same density as water, the 100.0 mL of HCl(aq) has a mass of 100.0 g. Qsurr = mc?T = (100.0 g)(4.184 J/(g·ºC)(25.5ºC - 19.0ºC) = (100.0 g)(4.184 J/g·ºC)(6.5ºC) = 2719.6 J - Qsurr = Qrxn -2719.6 J = Qrxn ?H = Q/n = - 2719.6 J/0.026796 mol = - 101492.8 J/mol = - 101.5 kJ/mol ± 0.5 kJ/mol Error: ?T: 0.5 + 0.5 = 1.0 ?H: 0.5 1.0 0.01 [ ( ¯¯¯¯ ) + ( ¯¯¯ ) + ( ¯¯¯¯¯¯¯¯ ) ][1.0] 100.0 6.5 0.026796 = 0.5 Trial 1 Table VI: Enthalpies of Reaction After Manipulating The Given

• Word count: 1262
• Level: International Baccalaureate
• Subject: Chemistry

#### Determining the Percent Yield of Calcium Carbonate

Determining the Percent Yield of Calcium Carbonate Purpose: To compare the theoretical amount to the actual amount of calcium carbonate and calculate its percent yield in the reaction between solutions of sodium carbonate and calcium chloride. Materials: stirring rods fine filter paper 2 small beakers 0.8 - 1.2 g of sodium carbonate Erlenmeyer flask electronic balance graduated cylinder safety glasses 2 g of calcium chloride deionized water funnel retort stand ring clamp Safety Procedures: Sodium carbonate and calcium chloride are both harmful if swallowed or inhaled and may cause irritation to skin, eyes and respiratory tract. Safety goggles, and proper attire are to be worn at all times during this experiment. In particular, no loose clothing should be worn and long hair should be tied back. Procedure: . The filter paper was measured for mass. 2. The clean, dry beakers were measured for mass. The sodium carbonate was added to one beaker and the calcium chloride was added to the other beaker. The beakers were measured for mass again. 3. About 25 mL of deionized water was added separately to each beaker. The solutions were stirred with the stirring rods until the solids were dissolved. 4. The calcium chloride solution was poured into the sodium carbonate solution. 5. The filter paper was adjusted on the funnel so as to completely

• Word count: 754
• Level: International Baccalaureate
• Subject: Chemistry

#### Chemistry Laboratory Report --- Evaporation: Intermolecular Attractions

Chemistry HL Chemistry Laboratory Report Evaporation: Intermolecular Attractions . Data Collected Table 1: Data collected by the class on different liquids Sample Physical Appearance Initial Temperature (?)±0.2 Final Temperature (?)±0.2 Change in Temperature (?) Polarity Molecular Formula Structural Formula Acetic Acid Clear liquid 25.2 24.6 -0.6 polar C2H4O2 CH3COOH Acetone Clear liquid 23.0 20.0 -3.0 polar C3H6O CH3COCH3 Aniline Grape Colour liquid 25.2 25.2 0.0 polar C6H7N (CH3)2C6H3NH2 Butanol Clear liquid 25.4 24.2 -1.2 polar C2H6O CH3CH2OH Butan-2-one Clear liquid 25.0 0.0 -15.0 nonpolar C4H10O CH3OHCH3 Benzaldehyde Clear liquid 22.0 22.0 0.0 polar C7H6O C6H5CHO Carbon Tetrachloride Clear liquid 25.0 9.2 -15.8 nonpolar CCl4 CCl4 Cyclohexane Clear liquid 27.0 0.0 -17.0 nonpolar C6H12 CH3CH2CH2CH2CH3 Cyclohexane Clear liquid 24.0 24.0 0.0 nonpolar C6H12 CH3CH2CH2CH2CH3 Chloroform Clear liquid 26.2 9.6 -16.6 polar CHCl3 CHCl3 Cyclohexanol Clear liquid 24.2 24.8 0.6 polar C6H11OH C6H11OH Di-ethyl Ether Clear Liquid 25 -12 -37.0 nonpolar C4H10O CH3CH2OCH2CH3 Ethyl Acetate Clear liquid 24.0 1.2 -12.8 nonpolar C4H8O2 CH3COOCH2CH3 Ethyl Ether Clear liquid 24.0 -9.0 -33.0 nonpolar C4H10O CH3CH2OCH2CH3 Ethyl Ether Clear liquid

• Word count: 930
• Level: International Baccalaureate
• Subject: Chemistry

#### Empirical Formula of Magnesium Oxide

Empirical Formula of Magnesium Oxide Name: Suleman Esam Date: 16/10-07 Purpose: The purpose of this lab was to experimentally determine the empirical formula of magnesium oxide. Background: Metal oxide is formed when a metal reacts with oxygen. Magnesium is an alkaline earth metal. When magnesium is heated, it reacts vigorously in the preset of oxygen. It is easy to ignite it when it is shred or powdered, but difficult when it is bulk. Once the metal is ignited, it is hard to extinguish it. Equipments: - Magnesium ribbon - Crucible & lid - Bunsen burner - Steel wool - Pipe clay triangle - Tripod - Crucible tongs - Balance - Goggles Procedure: . We took on our goggles and stated the experiment. 2. First we placed a tripod over a Bunsen burner. On the tripod, we placed a pipe clay triangle and over it the crucible & lid. 3. Then we lighted up the Bunsen burner; heating the crucible and the lid. We did that, because usually there is moist on things. And by steaming the moist, we will get a much better result and decrease the uncertainty. Then we turned off the Bunsen burner and left the crucible and its cover till it cooled to the room temperature. 4. While the crucible & the lid were cooling, we obtained a piece of magnesium ribbon and cleaned it with a steel wool. (There is a thin layer of oxide on the surface of magnesium. We cleaned it, so that

• Word count: 1047
• Level: International Baccalaureate
• Subject: Chemistry

#### Measuring the rate of reaction

MEASURING RATE OF THE REACTION OF YOUR CHOICE Research question: what is the rate of change of the reaction of decomposition of hydrogen peroxide using different amounts of different catalysts? Aim: to learn what effect do different amounts of biological catalyst liver and chemical catalyst MnO2 have on the rate of change of the reaction of decomposition of hydrogen peroxide. Hypothesis: we believe that the more catalyst is added, the faster the rate of reaction will be. It is also expected that MnO2 will be a better catalyst than the biological one i.e. liver. The equipment: The substances: Graduated tube water Clamp stand manganese oxide (MnO2) Water through 1M hydrogen peroxide (H2O2) Delivery tube liver Stopper Conical flask Stop watch The method: The reaction of decomposition of hydrogen peroxide is as follows: 2H2O2 (l) 2H2O(l) + O2(g) The rate of reaction will be calculated with reference to the amount of oxygen gases obtained. The amount of oxygen gases will be obtained using the process of displacement of water by O2 gases. The procedure is the following: . The apparatus (Picture 1.) is constructed for collecting the O2 gases. 2. The water through and the graduated tube are filled with water. 3. 10 ml of H2O2 is poured in the conical flask. 4. 0.2 g of MnO2 is added to

• Word count: 1581
• Level: International Baccalaureate
• Subject: Chemistry

#### Enthalpy. This experiment's aim is to find out, enthalpy change of a reaction, and to determine whether it is exothermic or endothermic.

Lab report on Emthapy Aim : This experiment's aim is to find out, enthalpy change of a reaction, and to determine whether it is exothermic or endothermic. Apparatus : Measuring Trays Weighing Scale Citric acid Sodium Carbonate Polystyrene cup Thermometer Reaction beaker + lid Copper Sulphate Zinc (powdered) Stopwatch Method: . Pour 25 ml of Copper Sulphate into the reaction beaker. 2. Place the reaction beaker in the polystyrene cup and put the thermometer in the cap. 3. Record the temperature of the solution. 4. Add the 0.5 grams of Zinc powder and close the lid immediately. 5. Record Maximum change of temperature. 6. Repeat steps 1 - 5 but this time use citric acid in place of Copper Sulphate and 3 grams of Sodium Carbonate instead of zinc. 7. Calculate energy change for the amounts of substance used. 8. Calculate the amount of the reactant in solution in the cup. 9. For each reaction calculate energy change per mole of the reactant which was in solution at the start. 0. Write a symbol equation for the first reaction and a word equation for the second one. 1. Write the energy change per mole of citric acid. 2. Draw an energy level diagram for the enthalpy change of both reactions. Diagram of Apparatus: Results: Starting temperature: Reaction Product Trial 1 Trial 2 Trial 3 Average temperature Zinc Sulphate 23 21 23 5 2 Sodium

• Word count: 629
• Level: International Baccalaureate
• Subject: Chemistry