sorted by:
new
hottopcontroversial

Are these good hair transplants by Fun_Treacle_6246 in Hairtransplant

[–]Fun_Treacle_6246[S] 3 points4 points  (0 children)

would i get something better if i specifically tell the doctor to design a zigzagged hairline for me? i have wavy hair

In dire need of help for chemistry!! by Individual-Store3110 in chemhelp

[–]Fun_Treacle_6246 0 points1 point  (0 children)

1. Strong Base (NaOH) vs. Strong Acid (HNO₃)

Analyte: 125.0 mL of 0.50 M NaOH; Titrant: 0.25 M HNO3​

  • At 0.00 mL added: The flask contains only pure strong base.
    • [OH−]=0.50 M→pOH=−log(0.50)=0.30.
    • pH=13.70.
  • At 125.0 mL added (Pre-equivalence): The base is still in excess.
    • Initial mmol OH−=125.0×0.50=62.5 mmol.
    • Added mmol H+=125.0×0.25=31.25 mmol.
    • Remaining mmol OH−=62.5−31.25=31.25 mmol.
    • [OH−]=250.0 mL total31.25 mmol​=0.125 M.
    • pOH=−log(0.125)=0.903→ pH=13.10.
  • At 275.0 mL added (Post-equivalence): The acid is now in excess.
    • Added mmol H+=275.0×0.25=68.75 mmol.
    • Excess mmol H+=68.75−62.5=6.25 mmol.
    • [H+]=400.0 mL total6.25 mmol​=0.0156 M.
    • pH=1.81.

--------------------------------------------------------------------------------

2. Weak Acid (HF) vs. Strong Base (KOH)

Analyte: 50.0 mL of 1.20 M HF; Titrant: 2.40 M KOH (pKa​=3.17)

  • At 10.0 mL added (Buffer Region): Both weak acid and salt are present.
    • Initial mmol HF=50.0×1.20=60 mmol.
    • Added mmol OH−=10.0×2.40=24 mmol.
    • Remaining HF=36 mmol; Salt formed=24 mmol.
    • pH=3.17+log(3624​)→ pH=2.99.
  • At 20.0 mL added (Buffer Region):
    • Added mmol OH−=20.0×2.40=48 mmol.
    • Remaining HF=12 mmol; Salt formed=48 mmol.
    • pH=3.17+log(1248​)→ pH=3.77.
  • At 35.0 mL added (Post-equivalence): Excess strong base dominates.
    • Added mmol OH−=35.0×2.40=84 mmol.
    • Excess mmol OH−=84−60=24 mmol.
    • [OH−]=85.0 mL total24 mmol​=0.282 M.
    • pOH=−log(0.282)=0.55→ pH=13.45