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.
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Answer:
tri-
Explanation:
Examples could be Tri-angle, Tri-cycle, Tri-ceratops
Answer:
<u>2-chlorohexane</u>
Explanation:
<u>In this figure</u> :
- There are 6 carbon atoms
- The Cl atom is bonded to the 2nd carbon atom
⇒ The Cl is a substituent group, termed as -chloro
⇒ Based on IUPAC nomenclature, the 6 atom chain starts with hex
⇒ There are only single bonds present, so it is an alkane
<u>The name is</u> :
Answer:
.259 g
Explanation:
PV = nRT
n = PV / RT
= .986 x 0.144 / .082 x 293.6
= .005897 moles
= .005897 x 44 g
= .259 g