Which of the following is a strong reducing agent? explain how you know.
a) mno4-
b) h+
c) h2
d) na
e) na+

The answer is: The principal idea here is how r they obtained: 
for example: sodium  u put equal molar concentrations of sodium (Na) and sodium ion (Na+)  together in a beaker , then dip in this solution a platinium wire (zero potential) which is connected to a normal hydrogen electrode (electrode with zero potential)  then u see the reading of the whole circuit 
if it is negtive, this means negative potential which means that the reducing property predominates where Na(reducing agent) is oxidized and electrons r accumulated on the platinum which gives it negative charge 
This means that (Na) is a reducing agent, its strength depends on the value of the potential obtained, and here the table can help you 
If u want to know if it's strong red. agent, look for it in the table, see if it has higher reduction potentail (or lower oxidation potential, same idea) than most other substances then it is reducing agent 
and vice versa 

So if we look at ur examples, u will find that MnO4- is the very strong oxidizing agent (has highest oxidation potential) (lowest reduction potential) 

H+ and H2 are both with zero potential, no redox properties 

And lastly Na and Na+: 
This u can know from ur knowledge in chemistry, that sodium is very rarely found in elemental form and always in the form of ion so u can deduce that Na is the very strong reducing agent 
or u can see the value of its standard oxi or red potetial and deduce which is the predominating form of them. 

I hope this helps

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer