Research Question: How does the reactivity of a Metal affect the enthalpy change for a reaction? Aim: To investigate how reactivity’s of various metals (Fe, Mg, Zn, Cu and Mn) affect the enthalpy change of the solution when it reacts with 25cm3 Copper II Sulphate. Theory: Specific heat capacity is the amount of energy needed to raise the temperature of one gram of a substance by 1 degree Celsius. Different materials absorb different amount of heat when their temperature rises by one degree Celsius. Therefore, every material has a different specific heat capacity. Metals usually heat up very easily because they have a low specific heat. Equations Mn(s) + CuSO4 (aq) MnSO4 (aq) + Cu (s) Fe(s) + CuSO4 (aq) …show more content…
Data Table 2: Table showing the average temperature (oC) (±0.5oC)per increment for the five metals reactions with Copper II Sulphate: Time (s) (±0.01s) Temperature (oC) (±0.5oC) Copper II (Cu) Manganese (Mn) Iron (Fe) Zinc (Zn) Magnesium (Mg) 0 25 25.3 25.0 25.0 25.0 30 25 25.3 30.7 25.0 25.0 60 25 64.3 33.7 32.7 64.7 90 25 75.7 35.3 34.0 92.7 120 25 78.3 36.7 35.3 91.0 150 25 77.7 37.0 36.0 88.3 180 25 76.3 37.0 36.7 86.7 210 25 74.3 39.0 38.0 83.3 240 25 72.3 39.7 38.3 81.3 270 25 70.7 40.0 39.3 79.0 300 25 69.0 40.7 40.0 78.0 330 25 41.0 40.7 77.0 360 25 41.0 43.3 390 25 41.7 45.0 420 25 42.0 47.0 450 25 42.0 48.3 480 25 42.0 48.7 510 25 410 49.0 540 50.0 570 49.3 600 49.0 Data …show more content…
Temperature (Magnesium) 25oC ±0.5 = 0.5/25 x 100= 2% Δ Temperature: (Tf – Ti) 0.5/98.82 x 100 = 0.5% 55.7 – 25 = 73.8 2% + 0.5% = 2.5% (of 73.8) = ±3.38oC Similarly evaluate other uncertainties By plugging in these values in the formula we can find energy released by the reaction: Q Magnesium= (25g) x (4.18) x (73.8) = 7714.2 Joules (J) = 7.714 Kilojoules (kJ) Absolute Uncertainty: Volume: 0.3/25 x 100 = 1.2% Energy uncertainty = 1.2% + 2.5% = 3.7% (of 7.71 kJ) = ±0.29 kJ Using the average mass of every metal used, we can find the number of mols through the formula n=
Then the mass of the copper metal and the percentage of Cu were obtained and compared throughout different groups and a mean and standard deviation was calculated for the
METHOD: The following procedure was taken from the 2017 Millsaps College lab manual.1 The experiment was split into two parts, part A and part B. Part A was to find the heat capacity while part B determined the specific heat of an unknown metal. This was the final goal of the lab. To start, a temperature probe had to be connected to a LabQuest2 data collection device. 100.0 mL of deionized had to be added into a Styrofoam cup.
Thermochemistry What is the specific heat of platinum if 1092 J of heat were released into a calorimeter when it was cooled by 65.2 C A 185 g sample of copper at 98.0 C was added to 102 g of water at 20.0 C in a calorimeter. The final temperature of the copper-water mixture was 31.2C. Calculate the specific heat of copper using this data. How much heat in kJ is required to raise the temperature of 250.0 g of Hg 52.0 C? the heat capacity of Hg is 0.14 J/gC.
2H3C6H5O7(s) + 3Mg(OH)2 (s) → Mg3(C6H5O7)2 (aq) + 6H2O (l) In order to get to this end point, we used Hess’ Law which says that the enthalpy of a net chemical reaction is the sum of the enthalpy changes of each individual step.
To better understand this law, Cu(s) was transformed with different reactions only to return back to Cu s). The initial and final mass of Cu(s) was recorded to give the percent recovery of copper product at
In addition, when both elements were carried out, it was noticeable that each of the test tubes feels warm. This indicated the reaction is an exothermic reaction because it produced heat. The pH level for magnesium chloride solution was neutral (not basic because of oxide layer) but basic for calcium chloride. It can be seen that calcium is more reactive than magnesium. This was because the lower the elements are down a group, the larger the size of its atomic radii.
Using the equation m = ΔTf/Kf , the molality of the unknown solution was found. Then, moles of unknown were calculated, which was used to calculate the average molar mass of unknown. Theory: After the experiment was completed, the data
The water percent is determined to be 42.06%. To find a percent error, a theoretical percent water must be used. To find the theoretical percent error divide the mass of water by the mass of magnesium sulfate heptahydrate and multiply by 100 to get a percent. The theoretical percent water is
75.33 grams Weight of the unknown = 0.23 grams Calculation : 75.33-75.10/0.36x100 = 63.8 % recovery Melting point of
The actual yield of the reaction was 4.411 grams of copper and was obtained through the experiment
Copper Cycle Lab Report Ameerah Alajmi Abstract: A specific amount of Copper will undergo several chemical reactions and then recovered as a solid copper. A and percent recovery will be calculated and sources of loss or gain will be determined. The percent recovery for this experiment was 20.46%.
Empirical Formula of Magnesium Oxide - Lab Report Background Information/Introduction: The aim of this lab is to determine the empirical formula of magnesium oxide by converting magnesium to magnesium oxide. As an alkali earth metal, magnesium reacts violently when heated with oxygen to produce magnesium oxide and magnesium nitride as a byproduct. In order to obtain only magnesium oxide, distilled water was added so that magnesium nitride will react and convert to magnesium hydroxide. Further heating then oxidizes all of the magnesium into magnesium oxide.
AUSTRALIAN PINE CONES-BASED ACTIVATED CARBON FOR ADSORPTION OF COPPER IN AQUEOUS SOLUTION MUSLIM A. Department of Chemical Engineering, Syiah Kuala Univeristy No. 7JalanTgk. Syech Abdul Rauf, Darussalam, Banda Aceh Indonesia
so when you multiply the 2.9 by 2 it gives you 5.8 cm3 which was my result. As the voltage increases the volume of hydrogen and oxygen increases. The standard deviation when I used 9 volts for hydrogen it was 0.7 + 0.23= 0.93 and 0.7-0.23= 0.47 so the range of values is between 0.47 cm3 to 0.93 cm3. When I used 9 volts for oxygen it was 0.3+0.12=0.42 and 0.3-0.12=0.18 so the range of values is between 0.18 cm3 to 0.42 cm3.
℃^(-1)×6.40℃±3.1 %=1337.6 J±4.06 % ∆H=(-1337.6 J±4.06 %) /(0.025 mol ±0.16 %)= -53504 J m〖ol〗^(-1)±4.22 % ∆H=-53504 J m〖ol〗^(-1)±4.22 %÷1000=-54 kJ m〖ol〗^(-1)±4.22 % Conclusion and