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Komok [63]
4 years ago
15

Identify the oxidizing agent and the reducing agent in the following reactions: (i) 8NH3( g) + 6NO2( g) => 7N2( g) + 12H2O( l

) (ii) Zn(s) +AgNO3(aq) => Zn(NO3) 2(aq) + Ag(s) (unbalanced)
Chemistry
1 answer:
shusha [124]4 years ago
4 0

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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weqwewe [10]

Answer:

The value of y = 5.1478

Explanation:

The linear equation is an equation obtained when a linear polynomial is equated to zero. When the solution obtained on solving the equation is substituted in the equation in place of the unknown, the equation gets satisfied.

The given equation: 5.3 x 10- (y)(2y) = 0

⇒ 53 - 2y² = 0

⇒ 2y² = 53

⇒ y² = 53 ÷ 2 = 26.5

⇒ y = √26.5 = 5.1478

8 0
4 years ago
What is the usable form of nitrogen that is found in the ground?
laila [671]
Nitrate

Hope this helps :))
5 0
3 years ago
What happens to dissolving rate when you increase KE?
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4 0
4 years ago
What is the molar concentration of 35 mL of H2SO4 that neutralizes 25 mL of 0.320M NaOH
stira [4]
V ( H2SO4) = 35 mL / 1000 => 0.035 L

M ( H2SO4) = ?

V ( NaOH ) = 25 mL / 1000 => 0.025 L 

M ( NaOH ) = 0.320 M

number of moles NaOH:

n = M x V

n = 0.025 x 0.320 => 0.008 moles of NaOH

Mole ratio:

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??

0.008 x 1 / 2 => 0.004 moles of H2SO4 :

Therefore:

M ( H2SO4) = n / V

M = 0.004 /  0.035

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hope this helps!



6 0
4 years ago
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer
lidiya [134]

The question is incomplete, the complete question is

Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V

Answer:

only IV and V

Explanation:

If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.

On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.

This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.

7 0
3 years ago
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