Kory goes outside to grill some hamburgers. He tells Ethan that when you cook meat, a chemical reation is taking place. Ethan sa
1 answer:
You might be interested in
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.
Learn more about Fuel Cell here : brainly.com/question/16612142
#SPJ4
Disclaimer : The question given was incomplete on portal, Here is the complete question.
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?
1) Balance the chemical equation.
2) List the known and unknown quantities.
Reactant 1: Ca(NO3)2.
Amount of substance: 3.40 mol.
Reactant 2: Li3PO4.
Amount of substance: 2.40 mol.
Product: Ca3(PO4)2
Mass: unknown.
3) Which is the limiting reactant?
<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.
<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.
4) Moles of Ca3(PO4)2 produced from the limiting reactant.
We have 3.40 mol Ca(NO3)2 of the limiting reactant.
The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.
5) Mass of Ca3(PO4)2 produced.
The molar mass of Ca3(PO4)2 is 310.1767 g/mol.
<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.
Option D.
.