6.8 is the pH of the solution after 10 ml of 5M NaOH is added.
Explanation:
Data given:
Molarity of C6H5CCOH = 0.100 M
molarity of ca(c6h5coo)2 = 0.2 M
Ka = 6.3 x 10^-5
first pH is calculated of the buffer solution
pH = pKa+ log 10 ![\frac{[A-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D)
pKa = -log10[Ka]
pka = -log[6.3 x10^-5]
pKa = 4.200
putting the values to know pH of the buffer
pH = 4.200 + log 10 
pH = 4.200 + 0.3
pH = 4.5 (when NaOH was not added, this is pH of buffer solution)
now the molarity of the solution is calculated after NaOH i.e Mbuffer is added
MbufferVbuffer = Mbase Vbase
putting the values in above equation:
Mbuffer = 
= 
= 0.01 M
molarity or [ A-] = 5M
pH = pKa+ log 10
pH = 4.200 + log 10 
pH = 4.200+ 2.69
pH = 6.8
Missing question:
(a) 1s2 2s2 2p6 3s1
(b) 1s2 2s2 2p6 3s2
(c) 1s2 2s2 2p6 3s2 3p1
(d) 1s2 2s2 2p6 3s2 3p4
(e) 1s2 2s2 2p6 3s2 3p5
Answer is: a) 1s²2s²2p⁶3s¹ (sodium).
Sodium have the largest second ionization energy, because when he lost one electron(first ionization energy), he have stable electron configuration of noble gas neon (1s²2s²2p⁶), so sodium do not need to lost second electron, because he will have unstable electron configuration.
Answer:
A GALAXY WIITH I YHINK MANY
Explanation:
8H⁺ + 5Fe²⁺ + MnO₄⁻ ⇒ Mn²⁺ + 5Fe³⁺ + 4H₂O
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According to the reaction, Fe</span>²⁺ and MnO4⁻<span> have following stoichiometric ratio:
n(</span>Fe²⁺) : n(MnO4⁻) = 5 : 1
n(MnO4⁻) = n(Fe²⁺) / 5
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So, for each mole of </span>Fe²⁺ it is needed 1/5 moles of MnO4⁻.<span>
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