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:
Even the most powerful light-focusing microscopes can't visualise single atoms. What makes an object visible is the way it deflects visible light waves. Atoms are so much smaller than the wavelength of visible light that the two don't really interact. To put it another way, atoms are invisible to light itself.
Explanation:
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Given :
Number of moles of CHF₃ is 1.7 .
Solution :
We know, 1 mole of any complex contains 6.022 × 10²³ molecules.
Let, 1.7 moles of CHF₃ contains n numbers of molecules.
So, n = 1.7 × 6.022 × 10²³ molecules
n = 10.2374 × 10²³ molecules
n = 1.0237 × 10²³ molecules
Hence, this is the required solution.
They loose a valence electron
First determine the formal oxidation numbers:
N changes from +2 to +5 going from NO to (NO3)- O remains -2 the whole time Cr changes from +6 to +3
Now write the half reactions, balance the oxygens with the required number of waters and then balance the hydrogens with the required number of protons:
Oxidation half reaction:
NO(aq) + 2 H2O(l) ---> (NO3)-(aq) + 4 H+(aq) + 3 e-
Reduction half reaction:
(Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l)
Now balance the number of electrons on both sides and add them together:
2 NO(aq) + 4 H2O(l) ---> 2 (NO3)-(aq) + 8 H+(aq) + 6 e- (Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l) --------------------------------------... 2 NO(aq) + (Cr2O7)2-(aq) + 6 H+(aq) ---> 2 (NO3)-(aq) + 2 Cr3+(aq) + 3 H2O(l)
Notice that the charge is the same in both sides, which is an indication that the redox equation has been balanced correctly:
-2 + 6 = -2 + 2(+3) +4 = +4
