Moles of 
,
Moles of acetic acid 
,
Mole fraction of water :
Therefore, mole fraction of water in this solution is 0.645 .
Hence, this is the required solution.
In calculating the relative atomic mass of an element with isotopes<span>, the relative mass and proportion of each is taken into account. For example, naturally occurring chlorine consists of atoms of relative isotopic masses 35 (75%) and 37 (25%). Its relative atomic mass is 35.5.</span>
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:
E = 3.035× 10-¹⁹J = 1.9eV
f = 4.58 × 10¹⁴Hz
Explanation:
wavelength = 6.55 × 10-⁷m
c = 3 × 10⁸m/s
f = ?
E = ?
a) f = c/wavelength
f = 3 × 10⁸/6.55 × 10-⁷
f = 4.58 × 10¹⁴Hz
b) E = hc/wavelength
E = 6.626×10-³⁴ × 3 × 10⁸/ 6.55 × 10-⁷
E = 3.035 × 10-¹⁹J
1ev = 1.6 × 10-¹⁹J
Therefore E = 3.035/1.6 = 1.9eV
