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Komok [63]
2 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]2 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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If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T
zhuklara [117]

Answer: The actual yield of B_2O_3 is 60.0 g

Explanation:-

The balanced chemical reaction :

B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)

Mass of B_2H_6 =Density\times Volume=1.131g/ml\times 36.9ml=41.7g

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles

According to stoichiometry:

1 mole of B_2H_6 gives = 1 mole of B_2O_3

1.51 moles of B_2H_6 gives =\frac{1}{1}\times 1.51=1.51 moles of B_2O_3

Theoretical yield of B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g

Percent yield of B_2O_3= 75.7\%

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

75.7\%=\frac{\text{Actual yield}}{79.3}\times 100

{\text{Actual yield}}=60.0g

Thus the actual yield of B_2O_3 is 60.0 g

7 0
2 years ago
Devon’s laboratory is out of material to make phosphate buffer. He is considering using sulfate to make a buffer instead. The pk
sesenic [268]

Answer:

Is not possible to make a buffer near of 7.

Optimal pH for sulfate‑based buffers is 2.

Explanation:

The dissociations of H₂SO₄ are:

H₂SO₄ ⇄ H⁺ + HSO₄⁻ pka₁ = -10

HSO₄⁻ ⇄ H⁺ + SO₄²⁻ pka₂ = 2.

The buffering capacity is pka±1. That means that for H₂SO₄ the buffering capacity is in pH's between <em>-11 and -9 and between 1 and 3</em>, having in mind that pH's<0 are not useful. For that reason, <em>is not possible to make a buffer near of 7.</em>

The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is <em>2.</em>

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I hope it helps!

4 0
2 years ago
A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca
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Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}     .....(1)

Molarity of BaCl_2 solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

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