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Brrunno [24]
3 years ago
12

Which ionic equation describes a redox reaction? A. Ag(+) + Cl- = AgCl B. 2H(+) + CO3(2-) = CO2 + H2O C. H(+) + OH(-) = H2O D. Z

n + Cu(2+) = Zn(2+) + Cu The marking scheme of the past paper this question is from says that the answer is D but how?
Chemistry
1 answer:
Gennadij [26K]3 years ago
6 0

<u>Answer:</u> The correct option is D. Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously. It is known as the reaction in which the exchange of electrons takes place.  

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

From the given ionic reactions:

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

<u>On the reactant side: </u>

Oxidation number of Zn = 0

Oxidation number of Cu = +2

<u>On the product side: </u>

Oxidation number of Cu = 0

Oxidation number of Zn = +2

As the oxidation number of Zn is increasing from 0 to +2. Thus, it is getting oxidized. Similarly, the oxidation number of Cu is decreasing from +2 to 0. Thus, it is getting reduced. Therefore, forming a redox couple

Hence, the correct option is D. Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the
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A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V

The mole fraction of nitrogen in the mixture is:

X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

Learn more: brainly.com/question/2060778

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