Lowest is Hydrogen highest is <span>Beryllium
-HOPE THIS HELPED </span>
Can you post a picture of that the checkings are
Answer:
When excess of carbon dioxide is passed in lime water, calcium carbonate is converted to calcium bicarbonate which is soluble, hence the milkiness due to calcium carbonate disappears.
Explanation:
Ca(OH)2+CO2 → CaCO3 (Milkiness) ↓+H2O
CaCO3+H2O+CO2 → Ca(HCO3)2 (soluble)
Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,
![Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)](https://tex.z-dn.net/?f=Cu%28s%29%2BPb%5E%7B2%2B%7D%28aq%29%5Crightarrow%20Cu%5E%7B2%2B%7D%28aq%29%2BPb%28s%29)
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
![E^0_{cell}=E^0_{cathode}-E^0_{anode}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7Bcathode%7D-E%5E0_%7Banode%7D)
![E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)
![E^0_{cell}=-0.13V-(0.34V)=-0.47V](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3D-0.13V-%280.34V%29%3D-0.47V)
Now we have to calculate the standard Gibbs free energy.
Formula used :
![\Delta G^o=-nFE^o_{cell}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-nFE%5Eo_%7Bcell%7D)
where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.
![\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-%282%5Ctimes%2096500%5Ctimes%20%28-0.47%29%29%3D%2B90710J%2Fmole%3D%2B90.71kJ%2Fmole%5Capprox%20%2B91kJ%2Fmole)
Therefore, the standard Gibbs free energy is +91 kJ/mole
I believe the answer is 3 - He hypothesized that all substances can be broken down into a smallest part called the atom.