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Artemon [7]
2 years ago
8

Consider these generic half-reactions. Half-reaction E° (V) X+(aq)+e−⟶X(s) 1.52 Y2+(aq)+2e−⟶Y(s) −1.17 Z3+(aq)+3e−⟶Z(s) 0.84 Ide

ntify the strongest oxidizing agent. X+ Y2+ X Z Y Z3+ Identify the weakest oxidizing agent. X Z3+ Y Y2+ Z X+ Identify the strongest reducing agent. Z3+ X+ Y2+ Z X Y Identify the weakest reducing agent. Y Z X+ X Y2+ Z3+ Which substances can oxidize Z ?
Chemistry
1 answer:
olya-2409 [2.1K]2 years ago
5 0

Answer:

     strongest oxidizing agent: X^{+}

     weakest oxidizing agent: Y^{2+}

     strongest reducing agent: Y

     weakest reducing agent: X

     X^{+} will oxidize Z

Explanation:

The higher the reduction potential of a species, higher will be the tendency to consume electrons from another species. Hence higher will be the oxidizing power of it's oxidized form and lower will be the reducing power of it's reduced form.

Alternatively, higher reduction potential value suggests that the oxidized form of the species acts as a stronger oxidizing agent and the reduced form of the species acts as a weaker reducing agent.

Order of reduction potential:

                       E_{X^{+}\mid X}^{0}(1.52V)> E_{Z^{3+}\mid Z}^{0}(0.84V)> E_{Y^{2+}\mid Y}^{0}(-1.17V)

So, strongest oxidizing agent: X^{+}

     weakest oxidizing agent: Y^{2+}

     strongest reducing agent: Y

     weakest reducing agent: X

As reduction potential of the half cell X^{+}\mid X is higher than the reduction potential of the half cell Z^{3+}\mid Z therefore X^{+} will oxidize Z into Z^{3+} and itself gets converted into X.

     

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2 years ago
Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,00
AnnZ [28]

Answer:

Hello some parts of your question is missing

Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.

Answers:

a) 46.666.66 g/mol

b) 20930.23 g/mol

c)43333.33 g/mol

Explanation:

A)The equal number of molecules of each sample can be calculated using  Mn = \frac{n(M1 + M2 + M3)}{3n}

because for the number of molecules to be equal : n1 = n2 = n3 = n

Mn = 46666.66 g/mol

B ) To calculate the equal masses of each sample

we apply this equation

Mn = \frac{W1 + W2 +W3}{\frac{W1}{M1} +\frac{W2}{M2}+ \frac{W3}{M3}  }

ATTACHED BELOW IS THE REMAINING PART OF THE SOLUTION

7 0
2 years ago
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Answer:

Sigma bonds: 10

Pi bonds: 4

Explanation:

The compound described must be CH2=CH-CO-CH≡CH. If we look at the compound closely, we will notice that there are 10 sigma bonds and 4 pi bonds.

There are three pi bonds between carbon atoms and one pi bond between a carbon and an oxygen atom (C=O). All these can easily be seen in the structure of the formula chosen in this answer.

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6 0
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Ethylene (C2H4) is the starting material for the preparation of polyethylene. Although typically made during the processing of p
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Answer:

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Explanation:

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We can establish the following relations:

  • 1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
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The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}

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