The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
<u>Answer:</u> The mass percent of hydrogen in methyl acetate is 8 %
<u>Explanation:</u>
The given chemical formula of methyl acetate is 
To calculate the mass percentage of hydrogen in methyl acetate, we use the equation:

Mass of hydrogen = (6 × 1) = 6 g
Mass of methyl acetate = [(3 × 12) + (6 × 1) + (2 × 16)] = 74 g
Putting values in above equation, we get:

Hence, the mass percent of hydrogen in methyl acetate is 8 %
The answer should be E,C,F,B,A,D in that order.
Answer:
It's not atom it is an element Strontium.