<u>Answer:</u> The temperature of the solution in Kelvins is 422.356 K
<u>Explanation:</u>
Temperature is defined as the measure of coldness or hotness of a body. It also determines the average kinetic energy of the particles in a body.
This term is expressed in degree Celsius, degree Fahrenheit and Kelvins. All these units are interchangeable.
The S.I unit of temperature is Kelvins.
We are given:
Temperature of a solution = 
Conversion used to convert degree Celsius and Kelvins is:
![T(K)=[273.15+T(^oC)]](https://tex.z-dn.net/?f=T%28K%29%3D%5B273.15%2BT%28%5EoC%29%5D)
Hence, the temperature of the solution in Kelvins is 422.356 K
Answer:
(i) Oxidizing Agent: NO2 / Reducing Agent NH3-
(ii) Oxidizing Agent AgNO3 / Reducing Agent Zn
Explanation:
(i) 8NH3( g) + 6NO2( g) => 7N2( g) + 12H2O( l)
In this reaction, both two reactants contain nitrogen with a different oxidation number and produce only one product which contains nitrogen with a unique oxidation state. So, nitrogen is oxidized and reduced in the same reaction.
Nitrogen Undergoes a change in oxidation state from 4+ in NO2 to 0 in N2. It is reduced because it gains electrons (decrease its oxidation state). NO2 is the oxidizing agent (electron acceptor).
Nitrogen Changes from an oxidation state of 3- in NH3 to 0 in N2. It is oxidized because it loses electrons (increase its oxidation state). NH3 is the reducing agent (electron donor)
(ii) Zn(s) +AgNO3(aq) => Zn(NO3)2(aq) + Ag(s)
Ag changes oxidation state from 1+ to 0 in Ag(s).
Ag is reduced because it gains electrons and for this reason and AgNO3 is the oxidizing agent (electron acceptor)
Zn Changes from an oxidation state of 0 in Zn(s) to 2+ in Zn(NO3)2. It is oxidized and for this reason Zn is the reducing agent (electron donor).
Balanced equation:
Zn(s) +2AgNO3(aq) => Zn(NO3)2(aq) + 2Ag(s)
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The new concentrations of 
and 
are 0.25M and 19M
Calculation of number of moles of each component,
Molarity of = number of moles/volume in lit = 0. 500 M
Number of moles = molarity of × volume in lit = 0. 500 M× 0.025 L
Number of moles of = 0.0125 mole
Molarity of = number of moles/volume in lit = 0. 38 M
Number of moles = molarity of × volume in lit = 0. 38 M× 0.025 L
Number of moles of = 0.95 mole
Calculation of new concentration at volume 50 ml ( 0.05L)
Molarity of = number of moles/volume in lit = 0.0125 mole/0.05L
Molarity of = 0.25M
Molarity of = number of moles/volume in lit = 0.95mole/0.05L
Molarity of = 19 M
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