0.0760 m
do this by:
finding the moles of NaOH which will be <span>5.702 E -3 m
</span>
next find the moles of H3PO4 which will be <span>1.90 E -3 m</span><span>
calulcate </span>25 ml sample molarity = 0.07603 m, just put 0.0760<span>
</span>
Answer: Reducing agent in the given reaction is 
.
Explanation:
A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.
In the given reaction, oxidation state of sulfur in is +2 and 
has 0 oxidation state.
In 
oxidation state of S is 2.5 and in 
oxidation state of I is -1.
Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.
Thus, we can conclude that reducing agent in the given reaction is .
The energy transformations are similar because they result into radiant energy.
As for the lamp, Electrical energy is transformed into light when the filament
or mercury vapor glows on passage of current.
The fire- chemical energy is turned to light energy during the combustion of carbon. Both products comprise of ultraviolet radiation which is a form of radiant energy.
Gray matter hope this helps!
