Theoretical yield of Al₂O₃: 1.50 mol.
<h3>Explanation</h3>
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How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?
Aluminum reacts to aluminum oxide at a two-to-one ratio.
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As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?
Oxygen reacts to produce aluminum oxide at a three-to-two ratio.
As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced?
Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.
I am guessing that your solutions of HCl and of NaOH have approximately the same concentrations. Then the equivalence point will occur at pH 7 near 25 mL NaOH.
The steps are already in the correct order.
1. Record the pH when you have added 0 mL of NaOH to your beaker containing 25 mL of HCl and 25 mL of deionized water.
2. Record the pH of your partially neutralized HCl solution when you have added 5.00 mL of NaOH from the buret.
3. Record the pH of your partially neutralized HCl solution when you have added 10.00 mL, 15.00 mL and 20.00 mL of NaOH.
4. Record the NaOH of your partially neutralized HCl solution when you have added 21.00 mL, 22.00 mL, 23.00 mL and 24.00 mL of NaOH.
5. Add NaOH one drop at a time until you reach a pH of 7.00, then record the volume of NaOH added from the buret ( at about 25 mL).
6. Record the pH of your basic HCl-NaOH solution when you have added 26.00 mL, 27.00 mL, 28.00 mL, 29.00 mL and 30.00 mL of NaOH.
7. Record the pH of your basic HCl-NaOH solution when you have added 35.00 mL, 40.00 mL, 45.00 mL and 50.00 mL of NaOH from your 50mL buret.
Answer:
Electrons
Explanation:
Cathode rays carry electronic currents through the tube. Electrons were first discovered as the constituents of cathode rays. J.J. Thomson used the cathode ray tube to determine that atoms had small negatively charged particles inside of them, which he called “electrons.”
1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
Pulmonary ventilation, and carbonic acid-bicarbonate buffering system
