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vladimir2022 [97]
3 years ago
15

PLEASE HELP!!!

Chemistry
1 answer:
Ksju [112]3 years ago
6 0

Answer : The correct option is, (B) Oxygen diffuses through the anode.

Explanation :

Hydrogen-oxygen fuel cell : As we know that the a fuel cell is a type of device that converts the chemical energy into electrical energy. In the hydrogen-oxygen fuel cell, the proton exchange membrane cell uses hydrogen and oxygen gas as a fuel.

In this cell reaction, hydrogen and oxygen gas are the reactants and water, electricity and heat are the products.

The half-cell electrode reaction are:

At anode : 2H_2(g)+4OH^-(aq)\rightarrow 4H_2O(l)+4e^-

At cathode : O_2(g)+2H_2O(l)+4e^-\rightarrow 4OH^-(aq)

That means the hydrogen undergoes reduction at anode and oxygen undergoes oxidation at cathode.

As per question, we can say that oxygen diffuses towards anode and hydrogen diffuses towards cathode.

Hence, the correct option is, (B) Oxygen diffuses through the anode.

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How many atoms are there in 5K2CO3
Dvinal [7]

Answer:

Depends on what are you refering to

Explanation:

So depending on what you are looking for (your question is quite vauge)

there are 5 atoms of the comopound (K2CO3)

within that compound, there are 2 atoms of Potassium and 1 atom of Carbonate. Within Carbonate there are 4 atoms (1 carbon and 3 oxygens)

so answers may be

5, 15, or 25.

I hope this helps.

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Is fluorine a nonmetal or metal?
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Read 2 more answers
Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con
arsen [322]

Answer:

a)  T_b=590.775k

b)  W_t=1.08*10^4J

d)  Q=3.778*10^4J

d)  \triangle V=4.058*10^4J

Explanation:

From the question we are told that:

Moles of N2 n=2.50

Atmospheric pressure P=100atm

Temperature t=20 \textdegree C

                      t = 20+273

                     t = 293k

Initial heat Q=1.36 * 10^4 J

a)

Generally the equation for change in temperature is mathematically given by

\triangle T=\frac{Q}{N*C_v}

  Where

  C_v=Heat\ Capacity \approx 20.76 J/mol/K

T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }

T_b-293k=297.775

T_b=590.775k

b)

Generally the equation for ideal gas is mathematically given by

 PV=nRT

For v double

 T_c=2*590.775k

 T_c=1181.55k

Therefore

PV=Wbc

Wbc=(2.20)(8.314)(1181_590.778)

Wbc=10805.7J

Total Work-done W_t

W_t=Wab+Wbc

W_t=0+1.08*10^4

W_t=1.08*10^4J

c)

Generally the equation for amount of heat added is mathematically given by

Q=nC_p\triangle T

Q=2.20*2907*(1181.55-590.775)\\

Q=3.778*10^4J

d)

Generally the equation for change in internal energy of the gas is mathematically given by

\triangle V=nC_v \triangle T

\triangle V=2.20*20.76*(1181.55-293)k

\triangle V=4.058*10^4J

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Has highest melting point of two bonds
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