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Alex777 [14]
2 years ago
10

What role do the nafion resin and h3po4 play? could you use concentrated hcl or h2so4 instead?

Chemistry
1 answer:
Anna [14]2 years ago
7 0
<h2>Answer</h2>

Nafion

Nafion is a sulfonated tetrafluoroethylene based fluoropolymer-copolymer discovered in the late 1960s by Walther Grot of DuPont. It is the first of a class of synthetic polymers with ionic properties which are called ionomers. Nafion's unique ionic properties are a result of incorporating perfluorovinyl ether groups terminated with sulfonate groups onto a tetrafluoroethylene (Teflon) backbone.Nafion has received a considerable amount of attention as a proton conductor for proton exchange membrane (PEM) fuel cells because of its excellent thermal and mechanical stability.

Role of Nafion

Nafion's properties make it suitable for a broad range of applications. Nafion has found use in fuel cells, electrochemical devices, chlor-alkali production, metal-ion recovery, water electrolysis, plating, surface treatment of metals, batteries, sensors, Donnan dialysis cells, drug release, gas drying or humidifaction, and superacid catalysis for the production of fine chemicals. Nafion is also often cited for theoretical potential (i.e., thus far untested) in a number of fields. With consideration of Nafion's wide functionality, only the most significant will be discussed below.

Role of Phosphoric acid in fuel cells

Phosphoric acid is used as an electrolyte in phosphoric acid fuel cells. Pure liquid phosphoric acid (H3PO4) saturated in a silicon carbide matrix (SiC) is used as electrolyte in different fuel cells. Operating range is about 150 to 210 °C. The electrodes are made of carbon paper coated with a finely dispersed platinum catalyst.

Use of H2SO4 and HCl

H2SO4 is normally used as electrolyte as Phosphoric acid. There is no evidences of use of HCl as electrolyte because HCl will produce Chlorine gas which is very reactive so HCl is avoided to use as electrolyte in fuel cells.


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Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that P_{1} = 100\,kPa, V_{1} = 500\,mL and V_{2} = 1000\,mL, then the new pressure of the gas is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 500\,kPa

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that V_{1} = 3.60\,L, P_{1} = 10\,kPa and P_{2} = 25\,kPa then the new volume of the gas is:

V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)

V_{2} = 1.44\,L

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that \frac{P_{2}}{P_{1}} = 3, then the volume ratio is:

\frac{V_{1}}{V_{2}} = 3

\frac{V_{2}}{V_{1}} = \frac{1}{3}

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

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Answer:

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