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Katena32 [7]
2 years ago
11

A student was comparing two samples with an equal number of carbon atoms. One sample contained only Carbon-12 atoms. One sample

contained only Carbon-14 atoms, which contain two more neutrons than Carbon-12 atoms. The student measured the mass of each sample and testing the reactivity of each sample.
Required:
What would best describe the results of the investigation?
Chemistry
1 answer:
Diano4ka-milaya [45]2 years ago
3 0

Answer:

3.14

Explanation:

A student was comparing two samples with an equal number of carbon atoms. One sample contained only Carbon-12 atoms. One sample contained only Carbon-14 atoms, which contain two more neutrons than Carbon-12 atoms. The student measured the mass of each sample and testing the reactivity of each sample.

Required:

What would best describe the results of the investigation?

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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

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   = (25 + 273 ) K

   = 298 K

<h3>What is Ideal Gas Law ?</h3>

It is expressed as

PV = nRT

Hence ,

n = \frac{PV}{RT}

   = \frac{(12)(3.5.10^{6} )}{(0.082057)(298)}

   = 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

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We know that

charge = moles of electron × F

          =  (3.435 ×10⁶ × 96500) C

           =  3.31 × 10¹¹ C

(c) We know that,

Voltage = \frac{energy}{charge}  =  \frac{J}{C}

1.44 V = \frac{energy}{3.31 . 10^{11} }

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.

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