4.
1 answer:
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Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:
To find,
Using 1 and 2 , we get:
<u>Size of the potassium ion = 1.33 Å</u>
If the coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.
(a) The reaction is
2H₂O + 2e⁻ → H₂ + 2OH⁻
According to the reaction, 2 moles of electrons flow per mole of reaction,
(b) Given
T = 25°C
Change Celsius into Kelvin
T = 25°C
= (25 + 273 ) K
= 298 K
<h3>What is Ideal Gas Law ?</h3>It is expressed as
PV = nRT
Hence ,
n =
=
= 1.7176 ×10⁶ mole
Now, From the given reaction we can say that ,1 mole of H₂ is produced by 2 moles of electron
Hence, 1.7176 × 10⁶ mole of H₂ is produced by
= 2 × 1.7176 × 10⁶ moles of electron
= 3.435 ×10⁶ moles of electron
We know that
charge = moles of electron × F
= (3.435 ×10⁶ × 96500) C
= 3.31 × 10¹¹ C
(c) We know that,
Voltage = =
1.44 V =
Energy = 1.44 V × 3.31 × 10¹¹ C
= 4.766 × 10¹¹ Joules
Thus from the above conclusion we can say that, coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.
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Question : Car manufacturers are developing engines that use H₂ as fuel. In Iceland, Sweden, and other parts of Scandinavia countries, where hydroelectric plants produce inexpensive electric power, the H₂ can be made industrially by the electrolysis of water.
a) How many moles of electrons flow per mole of reaction ?
b) How many coulombs are needed to produce 3.53X10⁶ L of H₂ gas at 12.0 atm and 25°C.
c) If the coulombs are supplied at 1.44 V, how many joules are produced?