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judyandaub1
16.10.2020 •
Chemistry
Polonium-191 decays by emitting an alpha particle. The daughter product also decays by alpha decay. What is the product of the two decays?
a.Hg-183
b.Hg-189
c.Po-197
d.Pb-193
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Ответ:
The correct option is a: Hg-183.
Explanation:
The alpha decay is given by:
Hence, if Polonium-191 decays by emitting an alpha particle we have:
Now, if Pb-187 also decays by emitting an alpha particle we have:
Therefore, the correct option is a: Hg-183.
I hope it helps you!
Ответ:
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.