Answer:
5.0 × 10²⁴ molecules
Explanation:
Step 1: Write the balanced double displacement reaction
2 NaOH + CuSO₄ ⇒ Na₂SO₄ + Cu(OH)₂
Step 2: Calculate the moles corresponding to 5.0 × 10²⁴ molecules of Na₂SO₄
We will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.
5.0 × 10²⁴ molecule × 1 mol/6.02 × 10²³ molecule = 8.3 mol
Step 3: Calculate the moles of CuSO₄ required to produce 8.3 moles of Na₂SO₄
The molar ratio of CuSO₄ to Na₂SO₄ is 1:1. The moles of CuSO₄ required are 1/1 × 8.3 mol = 8.3 mol.
Step 4: Calculate the molecules corresponding to 8.3 moles of CuSO₄
We will use Avogadro's number.
8.3 mol × 6.02 × 10²³ molecule/1 mol = 5.0 × 10²⁴ molecule
Answer:
hope it helps.
<h3>stay safe healthy and happy.<u>.</u><u>.</u></h3>
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.
