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Nana76 [90]
2 years ago
13

Commercial electrolysis is performed on both molten NaCl and aqueous NaCl solutions. Identify the anode product, cathode product

, species reduced, and species oxidized for the(a) molten electrolysis
Chemistry
1 answer:
olasank [31]2 years ago
5 0

The anode product is Cl2 (g) , the cathode product is Na (l) metal , the species that is getting reduced is Na+ (l) and the species that is getting oxidised is Cl- (l)

<h3>What is Electrolysis ?</h3>

Electrolysis is a process that involves the movement of ions to the oppositely charged cathodes.

For example: electrolysis of NaCl.

Electrolysis of NaCl : The commercial electrolysis of NaCl is performed on both molten NaCl and aqueous NaCl solutions.

<h3>What is Electrolysis of molten NaCl ?</h3>

1. When NaCl is heated at very high temperature ≈1070 K, it gets separated into ionised forms i.e. sodium ion and chloride ion.

The following reaction takes place:

NaCl (l)   →  Na+ (l) + Cl-(l)

2. During electrolysis the following reactions takes place at cathode and anode respectively:

at anode: 2Cl- (l)  →  Cl2(g) + 2e-

at cathode: 2Na+ (l) + e-   →   Na(l)

Overall reaction is :

2Na+(l) + 2Cl-(l)  →  2Na(l) +  Cl2 (g)

So, from the above reactions the following conclusions can be drawn :-

The product obtained at anode is Cl2 (g) and the product obtained at cathode is Na(l) metal. The species that is getting oxidised is Cl- (l) and the species that is getting reduced is Na+ (l) metal.

Learn more about the Electrolysis here: brainly.com/question/24063038

#SPJ4

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A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and
Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

    The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = 99.7^oC

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

1atm=760mmHg

or,

1mmHg=\frac{1}{760}atm

As, 1mmHg=\frac{1}{760}atm

So, 752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

K=273+^oC

As, K=273+^oC

So, K=273+99.7=372.7

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor  = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}

\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole

Thus, the mass of one mole of vapor is 32.0 g/mole

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Solution :

From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .

Moles of KBr in 102 g of potassium bromide.

n = 102/119.002

n = 0.86 mole.

So, number of miles of KCl produced are also 0.86 mole.

Mass of KCl produced :

m = 0.86 \times Molar \  mass \ of \  KCl\\\\m = 0.86 \times 74.5 \ gram \\\\m = 64.07\  gram

Hence, this is the required solution.

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Diano4ka-milaya [45]
To determine the mass of oxygen per gram of sulfur for sulfur dioxide, we simply obtain the ratio of the mass of oxygen and the mass of sulfur produced from the decomposition of sulfur dioxide. All other values given in the problem statement above are just to confuse us that the question is a difficult one. We do as follows:

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2 years ago
An atom with an atomic number of 10 and a mass number of 24 would have ________.
Anastaziya [24]

An atom with an atomic number of 10 and a mass number of 24 would have 14 neutrons.

<h3>What is atomic number and mass number?</h3>

The term atomic number, conventionally denoted by the symbol Z.

Atomic number indicates the number of protons present in the nucleus of an atom, which is also equal to the number of electrons in an uncharged atom.

The mass number of a given atom is defined by the number of protons and neutrons present in the nucleus. In simple,

Protons + Neutrons = Mass Number

To learn more about atoms, refer

https://brainly.in/question/5318

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6 0
1 year ago
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