<span>Ag+(aq) + e– → Ag(s) Eo = +0.80 V
Pb2+(aq) + 2e– → Pb(s) Eo = –0.13 V
=>
</span>
<span>2Ag+(aq) + 2e– → 2Ag(s) Eo = +0.80 V
Pb(s) → Pb2+(aq) + 2e– - Eo = – (-0.13 V) = +0.13 V
----------------------------------
2Ag(+) (aq) + Pb(s) -------> 2Ag(s) + Pb(2+)
</span>
<span>E cell = E0 cathode – Eo anode = 0.80V - (- 0.13V) = 0.80V + 0.13V = 0.93V.
Answer: + 0.93V
</span>
About 484 million miles
Fifth Planet from Our Star
Jupiter orbits about 484 million miles (778 million kilometers) or 5.2 Astronomical Units (AU) from our Sun (Earth is one AU from the Sun).
Explanation:
Im not sure what the other trait is so I couldn't do phenotype. I hope this still helps though
Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
