H2O = CO2 + 2 H2O Combination 4. AgBr(s) = Ag (s) + Br2(l) = 2 Ag + Br2 Combination 5. Mg(s) + H2SO4 (aq) = MgSO4 + H2 Decomposition Table Three: Formation of Salt via Double Replacement Reaction 0.1 KNO3 * (NO3)2 0.1 M AgNO3 0.1M NaCl Tube 1: KNO3+NaCl Tube 3: Fe(NO3)3+NaCl Tube 5: AgNO3+NaCl Prediction No change No change No change Observation Change:
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Materials 1. Sand 2. CuSO4 · xH2O salt 3. Evaporating dish 4. Analytical balance 5. Small testtubes 6. Drying flask 7. Thermometer 8. Electrical heater Experimental Procedure Results and calculations Mass of Test tube : 15.8945g Mass of Test tube + CuSO4 ∙ xH2O salt : 17.6698g Mass of Test tube + CuSO4 ∙ xH2O salt after heat for
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precipitation reaction using stoichiometry‚ accurately measure the reactants and products of the reaction‚ determine the actual yield vs. the theoretical yield and to calculate the percent yield. The equation that will be used is: Ba(NO3)2 (aq) + CuSO4 (aq) → BaSO4 (s) + Cu(NO3)2 (aq) Method 1. Gather materials needed for experiment which included: a. Small test tube with lip b. Large beaker c. Small graduated cylinder d. Large graduated cylinder e. One 9in
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January 4‚ 2013 Chemistry Honors Title: Hydrates Purpose: To find the mass percentage of water in the hydrate. Materials: * Watch glass * 2.82 g hydrated crystal (CuSO4*5H2O) * Evaporating dish * Bunsen burner * Electronic balance * Metal tongs * Ring stand Procedure: 1) Determine the mass of evaporating dish and watch glass. 2) Add between 2 and 3 grams of the hydrated crystal to the evaporating dish. 3) Determine the mass of the dish and crystal
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We were given KOH and KNO3 as our salts. We weighed the calorimeter and water then subtracted accordingly o get the mass of the calorimeter and that of the water. The lab instructed on the amounts of water and salt to be used. The initial temperature of the water in the KOH reaction was 22.8 degrees Celsius and the final temperature of the mixture determined from the graph was 37.0 degrees Celsius. The graph of the KOh showed it to be an exothermic salt and the KNO3 to be and endothermic salt
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Data Collection and Processing Aspect One -2.00grams of Sr(NO3)2 used -2.00grams of CuSO4 -Reactants: Sr(NO3)2 and CuSO4 Trial Mass of Beaker (g) Mass of Beaker with Sr(NO3)2 (g) Mass of Beaker (g) Mass of Beaker and CuSO4 (g) Mass of Filter Paper (g) Mass of Filter Paper and Contents (g) 1 111.08±0.01 113.08±0.01 111.1±0.01 113.1±0.01 1.28±0.01 2.93±0.01 2 111.1±0.01 113.1±0.01 111.23±0.01 113.23±0.01 1.27±0.01 2.98±0.01 3 111.26±0.01 113.26±0.01 111.09±0.01 113.09±0.01 1.27±0.01
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Before we heated the blue CuSo4 Solution‚ there were still some crystals remain in the bottom. But when we finished heat the solution‚ the crystals were completely dissolved. (CuSo4 solution was blue) While we added 1.49 gram of iron filings (black) in to the breaker in small increments. But there were some Iron filings left on the weight cup. In the first
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of a unknown solution: Beers Law. We determined the concentration of a unknown CuSO4 solution by measuring its absorbance with the colorimeter. With all the calculations we were able to solve the linear regression Equation of absorbance vs. concentration and the alternate method. Materials Vernier LabPro or CBL 2 interface .40 M CuSO4 solution Computer or handheld CuSO4 unknown solution Vernier Colorimeter Pipet
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Using tongs‚ take the crucible out of clay triangle until it cools. 6) Carefully pour CuSO4 from crucible to massing tin. 7) Record the mass of massing tin with CuSO4 inside. 8) Clean up lab station. Data: Table one displays our data for percent of water in bluestone. | |Massing Tin |Massing tin with |Crucible with CuSO4 |Bluestone without crucible| CuSO4 | | | |bluestone |
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the mass of the hydrate and crucible gives the mass of the CuSO4 hydrate. Mass of hydrate = (mass of crucible + hydrate) - (mass of empty crucible) Mass of hydrate = (20.000g) – (15.000g) Mass of hydrate = 5.000g 2. Heat the hydrate. After heating‚ record the mass of the crucible and the dehydrated compound. No calculations for this step 3. Subtracting the mass of the empty crucible from the mass of the crucible and dehydrated CuSO4 gives you the mass of the dehydrated compound. Mass of dehydrate
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