Answer:
Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq)
1.35 V
Explanation:
Given that the reduction potential of
Cu^2+(aq) + e- -----> Cu^+(aq) is +0.15 V
While the reduction potential of
Au3+(aq) + 3 e- -----> Au(s) is + 1.50V
It is clear that the Cu^+(aq)/Cu^2+(aq) system is the anode while Au^3+(aq)/Au(s) system is the cathode based on the reduction potentials shown above. The number of electrons transferred (n) =3
E°cell = E°cathode - E°anode
E°cell= 1.50-0.15
E°cell= 1.35 V
A volatile impurity, is an impurity that evaporates quickly, so, if you were to conduct an experiment on a substance with this type of impurity, the evaporation point, and the mass of the substance at high temperatures would be in error.
<span>If it is known that at 60 °C and 745 torr UF₆ is present only as a gas then we can make the assumption that:
</span>(n/V) = P/RT = (745 torr)/(62.36 L-torr/mol•K)(337 K) = 0.0354 mol/L
Therefore the density will be:
<span>
Density = (352.02 g/mol)(0.0354 mol/L)/(1000 cm³/L) = 0.0125 g/cm³
</span>
I believe the answer would be D.
You get 42 when you multiply 6 & 7.
