1. Obtain 10 ml of 0.75 M Cu(NO)h and pour it into a 250 mL beaker. Add 90 mL of
ID: 547368 • Letter: 1
Question
1. Obtain 10 ml of 0.75 M Cu(NO)h and pour it into a 250 mL beaker. Add 90 mL of water to the beaker as well. Slowly add 30 mL of 3 M NaOH to the solution in the beaker, use a glass rod and gently stir the resulting mixture. Describe the reaction in your lab notebook 2. Add two to three boiling chips and place the beaker on a hotplate. Heat the solution until it is gently boiling. Describe the reaction in your lab notebook. 3. Allow the black CuO to settle, then carefully pour off the liquid. Add 100 ml of hot distilled water, stir, and then pour off the water. Why are you washing the solid with hot water? What are you removing? 4. 5. Add 15 ml of 6.0 M H:SO.. What copper compound is present in the beaker now? Add 2.0 g of zinc metal all at once and stir until the solution is coloriess. Describe the reaction in your lab notebook. 6. When the evolution of gas has slowed. Heat the solution gently (do not boil) and then allow it to cool. What gas is formed during the reaction? 7. When the gas evolution has ceased, pour of the solutions and transfer the copper precipitate to a pre-weighed evaporating dish. Wash the copper with water and then with methanol. Allow the methanol to evaporate and then weigh your product. 8.Explanation / Answer
Q2
V = 5 mL of Cu(NO3)2
M = 0.75 M
mol of Cu(NO3)2 = MV = 0.75*5*10^-3 =0.00375 mol of Cu2+
we could theoretically form 0.00375 mol of Cu(s)
mass = mol*Mw = 0.00375*63.5 = 0.238 g of Cu(s)
best answer is 0.24 g of Cu(s)
Q9
sources of error:
true, not heating enough will not let fumes to form completely and preicpitate to form
not drygin copper will have H2O present as hydrate
not adding Zn will clearly not form H2(g)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.