Answer:
Potential energy for n = 6 Bohr orbit electron is -1.21*10⁻¹⁹J
Explanation:
As per the Bohr model, the potential energy of electron in an nth orbit is given as:
here:
k = Coulomb's constant = 9*10⁹ Nm2/C2
Z = nuclear charge
e = electron charge = 1.6*10⁻¹⁹ C
r(n) = radius of the nth orbit = n²(5.29*10⁻¹¹m)
Substituting for k, Z(= 1), e and r(n) in the above equation gives:
Cell reaction overall is Zn(s)+2H+(aq)→Zn²+(aq)+H2(g).
The half-reaction of oxidation is Zn(s)→Zn²+(aq)+2E- and E°zn2+/zn=0.76v
Half reaction reduction 2H+(aq)+2∈-→H2(g) and E°H+/H2+=0.00v
Cell potential is E°cell=E°cathode-E°anode
=E°H+/H2e-E°zn2+/zn
=0.00v-(-0.76v)
=0.76v
Nernst equation
Ecell = -0.059W/N log [zn²+]PH2/[Zn][H+]²
Ecell = 0.66v
[zn²+]=1.0M
1=[Zn]
PH2=1atm
[H+]=?
n = number of moles of electrons transfered in the cell=2mol
Ecell=E°cell -0.059W/n log [Zn²+]/[Zn][H+]²
0.66v = 0.76 - 0.059W/2 log 1.0×1.0/1.0×[H+]²
0.059W/2 log 1/[H+]² = 0.76v-0.66v = 0.10v
log 1/[H+]² =0.10v×2/0.059W =3.4
-2log [H+] = 3.4
log [H+] = 1.7
[H+] =10-n
=0.020
=2.0×10-²M
The concenrtation in cathodic compartments is 2.0×10-²M
Answer:
The answer to your question is the letter C. 2.906 x 10²³ atoms
Explanation:
Data
mass of Lead = 100 g
The Number of atoms = ?
Process
1.- Look for the atomic number of Lead in the periodic table
Atomic number = 207.2 g
2.- Use proportions and the Avogadro's number to calculate the number of atoms.
207.2 g ------------------ 6.023 x 10²³ atoms
100 g ----------------- x
x = (100 x 6.023 x 10²³) / 207.2
x = 6.025 x 10²⁵ / 207.2
x = 2.906 x 10²³ atoms
Answer:
Either B or C. Composition or the Distance from the Earth.
