Answer:
Paper is Not Reuseable like plastic.
Explanation:
<span>It is known as the axon</span>
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)
<em>K</em> = 2.4 × 10^(-72)
<em>Step 1</em>. Determine the <em>value of n
</em>
Zn^(2+) + 2e^(-) → Zn
2Cl^(-) → Cl_2 + 2e^(-)
Zn^(2+) + 2Cl^(-) → Zn + Cl_2
∴ <em>n</em> = 2
<em>Step 2</em>. Calculate <em>K</em>
log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62
<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)
