First, we had to calculate how many grams of copper (II) sulfate we needed to form 100 mL of a 0.200 M solution of copper (II) sulfate. We determined that we needed to use 4.994g of copper (II) sulfate to make the solution. We added distilled water to the 4.994g of copper (II) sulfate in a beaker until it reached 100 mL. Then we put the beaker on a hot plate and added a magnetic stirrer. We determined that the mass of zinc necessary to completely react with the copper (II) ions in the solution was 1.308g.
CHEMISTRY Name _____________________________ Lab - Molar Volume of a gas lAB 12 OBJECTIVE OF THE LAB: The purpose of this lab is to determine the molar volume of a sample of Hydrogen gas that is created through the reaction of Magnesium and Hydrochloric Acid. The concept of molar volume is that 1 mole of a gas occupies 22.4 liters at STP. Unfortunately the conditions of the lab are not at STP. You will have to use gas law formulas to calculate the volume that one mole of this gas would occupy at STP. In addition, since this lab is being done over water, and water will evaporate at any temperature, the vapor pressure of water must be determined.
Do not ingest chemicals. Have all plans for experiments approved by the instructor. Wash hands after the experiment. Procedure The following equipment is available for your experiment. magnets zipper locking plastic bags beakers Petri dishes micro spatulas tongs 13- x 100-mm test tubes; rubber or cork stoppers to fit rings, ring stands, gauzes, burners, ignitors ice paper towels filter paper funnels Using this equipment, devise a procedure for the separation of the four substances.
Question: How can you determine that type of displacement reaction that is occurring? Materials: • 3 test tubes • 3 rubber stoppers • 15 cm of magnesium ribbon • Copper(II) sulfate crystals, fine • Calcium chloride • Sodium carbonate • 30 mL of warm water Safety Precautions: • Wear safety goggles , gloves, and a lab apron • Avoid skin contact with copper (II) sulfate dust and solution • Keep work area clean wipe up any spills and inform your teacher immediately Procedure: • Refer to page 210 and 211 in On Science 10 Observations: Before: • Solid • Blue crystals • Small • Shiny • solid • White • A bit shiny • Large crystals • White • Powdery • Soft After • liquid • The substance formed a bluish-green
1 mol of Zn Multiplied by the mole ratio of Zn to Cu(A) 2 mol of Cu(A) in the reaction. 65.39 grams of Zn Multiplied by the molar mass of Zn 1 mol of Zn = 0.1918 grams Zinc Any amount of Zinc that is 0.1918 grams or less makes Zinc the limiting reactant in the reaction of Cuprous (Cu+1). I will use 0.70* grams of Zinc. The ratio of moles
The solution starts to turns into a black color. Observation 4 Zinc Sulfate was placed into the test tube Green bubbles starts to come out. Observation 5 Hydrochloric Acid was added It turned clear and into a liquid solution. Observation 6 Zinc was added Silver, rock lie, and hard. Observation 7 Hydrogen Gas was being performed.
Year 11 Chemistry Electrochemical Cells Purpose To determine the order of reactivity for the metals copper, iron, lead and zinc using the least amount of electrochemical cells as possible. Materials List • Four 100ml beakers • 4 strips of filter paper approximately 15cm in length • Voltmeter • Sodium Chloride solution at 10% concentration • Zinc, Copper, Lead and Iron metal strips (approximately 10 cm) • 50 ml of Copper sulphate, Zinc sulphate, Lead nitrate and Iron nitrate solutions • Scissors Procedure - Place the 50ml solution of zinc sulphate in a 100ml beaker, and 50ml solution of copper sulphate in a different 100ml beaker. Place the zinc metal strip in the zinc sulphate beaker and the copper metal strip with the copper sulphate beaker.
Zn + I2 ⟶ ZnI2 | BaI2·2H2O +ZnSO4·7H2O⟶ZnI2+BaSO4+9H2O | Zinc used: (2g x $0.0625)=$0.125Iodine used:(3g x $0.1003)=$0.301Material cost for 3.1g of ZnI2 =$0.498Material cost for 10g of ZnI2 = $1.6 | 0.48g of Zinc Sulfate heptahydrate used= $0.01940.62g of Barium Iodide dehydrate used= $0.549Material cost for 0.48g of Zinc Iodide= $0.5684Material cost for 10g of ZnI2 = $11.84 | % Yield from Zn and I2 reaction: 98.6%% Error: 1.39% | % Yield from BaI2·2H2O and ZnSO4·7H2O reaction:103.6%% Error: 3.6% | The data above shows that Zinc Iodide formation through elements is much more efficient than using compounds. It costs less than half amount for Zn and I2 and the percent yield is also much closer to 100 when we use zinc granules and iodine crystals. Through the double replacement reaction, we gain a solid precipitate of Barium Sulfate at the end and the zinc iodide is the supernatant that we heated to get in solid form. It was not a completely pure method since the easier method is possible to form zinc iodide using only zinc and iodine. It was time consuming in terms of decanting the zinc iodide from barium sulfate solid.
Materials: copper wire AgNO3 solution Sandpaper Stirring Rod 50-mL Graduated Cylinder 50-mL Beaker Funnel Filter Paper 250-mL Erlenmeyer Flask Ring Stand Small Iron Ring Plastic Petri Dish Paper Clip Bunsen Burner Tongs Pre-Lab Questions: 1.) Read the entire CHEMLAB. 2. )Prepare all written material that you will take