Atomic mass iron ( Fe ) = 55.84 a.m.u
55.84 g ------------ 6.02x10²³ atoms
24.0 g ------------- ??
24.0 x ( 6.02x10²³) / 55.84
=> 2.58x10²³ atoms
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Answer:
Ecell = +0.25V
Explanation:
the half-cell reactions for a voltanic cell
cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)
anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻
we have the standard cell potential E⁺cell = 0.18V at 80C respectively
Q = [H⁺]/[Cl⁻]
sub for [H+] = 0.10M and [Cl-] = 1.5M
Q= 0.1M/1.5M
Q = 0.067
Ecell = E⁺cell - 
logQ
= 0.18 - 
log 0.067
0.18- 0.059(-1.174)
Ecell = +0.25V
The answer is letter A definitively .
the molar mass of the element is 81.36 g/mol
<u><em>calculation</em></u>
step 1 : multiply each %abundance of the isotope by its mass number
that is 79.95 x 29.9 =2391
81.95 x 70.1 = 5745
Step 2: add them together
2390.5 + 5744.7 =8136
Step 3: divide by 100
= 8136/100 = 81.36 g/mol
