Since that E°cell =E°cathode - E°anode then
0.62V = 0.34 - E°anode
Therefore E° anode is = -0.28V
Answer is: 18 moles of lead(II)-nitrate.
Balanced chemical reaction:
3Pb(NO₃)₂ + 2AlCl₃ → 2Al(NO₃)₃ + 3PbCl₂.
n(Al(NO₃)₃) = 12 mol.
From chemical reaction: nAl(NO₃)₃) : n(Pb(NO₃)₂) = 2 : 3.
n(Pb(NO₃)₂) = 3 · 12 mol ÷ 2.
n(Pb(NO₃)₂) = 18 mol; amount of substance.
Al(NO₃)₃ is aluminium nitrate.
AlCl₃ is aluminium chloride.
Answer:
1.345*10⁻⁴ mol/L
15.023 mg/L
Explanation:
The chemical <u>formula of ferrocene</u> is Fe(C₅H₅)₂, thus its molecular weight is:
55.845 g/mol + 10*12g/mol + 10 *1g/mol = 185.845 g/mol
- The moles of Fe contained in 25 mg (or 0.025 g) of ferrocene are:

- The final volume is 500 mL, or 0.5 L. So the iron concentration in mol/L is:

- We can convert that value into mg/L:

Answer:
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Explanation:
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