Goal: To make 1 grams of Cu and 2.54 grams of ZnSO4 Anhydrous
Equation: Zn + CuSO4.5H2O ------> ZnSO4+ Cu
Procedure:
First, we measured out 1.03 grams of Zn and 3.93g grams of CuSO4. Next we took the two and mixed them in a flask together. After we put them in the flask together we added enough water to get the reaction started. We swirled in around to make sure all the Zn reacted with the SO4. After the reaction took place, we let the flask sit for awhile to let all the copper separate from the ZnSO4. After the cooper settle to the bottom, we weighted our filter paper and the beaker that we are going to put the zinc sulfate. After we weighed the materials for the next step, we took the filter and fitted it into the beaker. After the paper was in the beaker my group ported the solution in the
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The first major set back we had is we had to restart twice because the first time we didn’t measure out the amounts correctly and the second time is because we used tap water in the reaction instead of distilled water. Another major problem we had was our copper was a mixture of copper and copper(ll). This was really hard to find the amount if each type of copper in the solution.
Actual Calculations: We were planning on getting 1.00g of Cu and 2.54g of ZnSO4 anhydrous crystal. What we ended up getting were 2.54g of ZnSO4 so exactly the right amount, but for the Cu we didn’t quite get what we were after. The final weight of our Cu was 1.12g we calculated out that if we had 0.0157 moles of Cu we would have ended up with 1.00g of Cu we got 1.12g so what we suspect to have happened is that we got 48% 2CuO which with 0.0157 moles of it we would have had 1.25g of it. By the look of our copper we can assume we got 48% 2CuO which should have resulted in 0.60g of it. Pairing that with a 52% Cu we should have gotten 0.52g of it added together they equal 1.12g that is exactly we got in the end.
Compress the safety bulb, hold it firmly against the end of the pipette. Then release the bulb and allow it to draw the liquid into the pipette.
Experiment: First prepared a well plate with the appropriate amounts of distilled water, HCl, and Na2S2O3 in each well according to the lab manual. The well where the reaction
Repeat for each trial. Rinse volumetric pipette with vinegar and drain into the waste beaker. Weigh and record the mass of each 200mL beaker. Add 10.00mL of vinegar into each beaker and weigh them and record their again. Add 50mL of de-ionized water to the beakers and place them under the drop counter on top of a stir plate, submerging the pH meter into the solution. Place the stir bar into the beaker and carefully turn on the stir plate so that the stir bar spins without splashing or hitting the sides of the beaker or the pH
· I made sure I put the same amount of water in the copper can.
This graph shows that as enzyme concentration increases absorption also increases. In this case absorbance can be used to measure the enzyme’s activity, the higher the absorption the higher the activity. Since absorption increases as enzyme concentration increases, enzyme activity is promoted by increased enzyme concentrations. After a certain point enzyme activity would fail to increase as a result of increased enzyme concentration since there wouldn’t be enough substrate for all of the enzymes to react with.
Introduction: The purpose of this lab was to cycle solid copper through a series of chemical forms and return it to its original form. A specific quantity of copper undergoes many types of reactions and goes through its whole cycle, then returns to its solid copper to be weighted. We observed 5 chemical reactions involving copper which are: Redox reaction (which includes all chemical reactions in which atoms have their oxidation state changed), double displacement reaction, precipitation reaction, decomposition reaction, and single displacement reaction. 4HNO3(aq) + Cu(s) --> Cu (NO3)2(aq) + 2H2O (l) + 2NO2(g) Oxidation reduction reaction Cu (NO3)2(aq) + 2 NaOH (aq) --> Cu (OH)2(s) + 2 NaNO3(aq) Precipitation Reaction Cu (OH)2(s) + heat --> CuO (s) + H2O (l) Decomposition reaction CuO (s) + H2SO Data Results: (mass of copper recovered / initial mass of copper) x 100 Mass of copper recovered: 0.21 Initial mass of copper: 0.52 (0.21/0.52)x100 =40.38%.
According to the graph on amylase activity at various enzyme concentration (graph 1), the increase of enzyme dilution results in a slower decrease of amylose percentage. Looking at the graph, the amylose percentage decreases at a fast rate with the undiluted enzyme. However, the enzyme dilution with a concentration of 1:3 decreased at a slow rate over time. Additionally, the higher the enzyme dilution, the higher the amylose percentage. For example, in the graph it can be seen that the enzyme dilution with a 1:9 concentration increased over time. However, there is a drastic increase after four minutes, but this is most likely a result of the error that was encountered during the experiment. The undiluted enzyme and the enzyme dilution had a low amylose percentage because there was high enzyme activity. Also, there was an increase in amylose percentage with the enzyme dilution with a 1: 9 concentrations because there was low enzyme activity.
The purpose of this lab is to determine the empirical formula of copper oxide (CuxOy) through a single-displacement reaction that extracts the copper (Cu) from the original compound. In order to do this, hydrochloric acid (HCl) was mixed in with solid CuxOy; the mixture was stirred until the CuxOy was totally dissolved in the solvent. Zinc (Zn) was then added to the solution as a way to enact a single displacement reaction in which Cu begin to form on the Zn; the Cu gets knocked off the Zn through gentle stirring. To isolate the Cu, the supernatant liquid was decanted and the Cu was then washed with first water then second, isopropyl alcohol. Once done, the hydrated Cu is transferred onto an evaporating dish where it was heated multiple times
In the lab the reaction that took place was a synthesis reaction. A synthesis reaction, is a type of chemical reaction in which two or more simple substances combine to form a new product. The reactants may be elements or compounds. In this case it is a gas and a metal that will react and produce a compound. The general form of a synthesis reaction is, A + B → AB. In order for this lab to be done successful you need knowledge on, percent composition, the empirical and molecular formula, the law of conservation of mass, moles and molar mass, qualitative and quantitative. To begin, the percent composition of a compound is the percent of the total mass that each element has in that compound. Every compound would have a certain percent composition. To calculate percent composition of a compound, you would have to determine the total molecular mass of the compound. For example, for H2O the total molar mass would be 18.00g/mol. You would then input the mass of one of the elements and the molar pass into the equation % by weight (mass) of element = (total mass of element present ÷ total mass of compound) x 100 to find out the percent composition. So for Oxygen it would be, % of O = (16.00g ÷ 18.00g/mol) x 100 which would equal 88.9%. Therefore the percent composition of O in this compound is roughly 88.9%. Furthermore, the molecular formula is the number and types of atoms that are existing in a single molecule of a substance. The empirical formula also known as the simplest formula is the ratio of elements present in the compound. The key difference between these two is that the empirical formula shows the simplest positive integer ratio of atoms of each element present in a compound whereas the molecular formula of a compound is a way ...
== == I completed a table to show my results, here is the table: Table 1. Results of different changes of substances Part A Copper (II) Sulfate and Water Reactant description Water (reactant): Color: Colorless Transparency:
TIME - 1 minute. The longer the ions have to move, the more copper is
To investigate the temperature change in a displacement reaction between Copper Sulphate Solution and Zinc Powder
81/125 = 0.648g To calculate how much zinc oxide is produced from 2g of calamine 0.648 is doubled: 0.648 x 2 = 1.296g To calculate how much zinc oxide is produced from 3g of calamine 0.648 is tripled and so on when increasing the mass of calamine. On the next page there is a table of the theoretical conversions for how much zinc oxide is produced from using certain amounts of calamine (1g-9g) Mass Of ZnCO₃ (g) Mass of ZnO Produced (g) 1 0.648 2 1.296 3 1.944 4 2.592 5 3.24 6 3.888 7 4.536 8 5.184 9 5.832 Here is a graph of the results predicted from the table………… This experiment is going to see how much zinc oxide can be obtained from calamine. It shall also show how close to the conversion can be achieved in practice. Hypothesis In a previous experiment where copper carbonate was obtained from malachite the results showed that as the more malachite was used the more product was produced.
of Copper Sulphate. To do this I plan to work out the amount of water
The Electrolysis of Copper Sulphate Aim Analyse and evaluate the quantity of Copper (Cu) metal deposited during the electrolysis of Copper Sulphate solution (CuSo4) using Copper electrodes, when certain variables were changed. Results Voltage across Concentration of solution electrode 0.5M 1.0M 2.0M 2 5.0 10.6 19.5 4 10.5 19.8 40.3 6 14.3 26.0 60.2 8 15.2 40.4 80.3 10 15.0 40.2 99.6 12 15.1 40.0 117.0 Analysing/Conclusion The input variables in this experiment are; concentration of the solution and the voltage across the electrodes. The outcome is the amount of copper gained (measured in grams) at the electrodes. By analyzing the graph, we can see the rapid increase of weight gained for the 2.0 molar concentration as the gradient is steeper.