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Olenka [21]
3 years ago
15

A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small bin

ding energy per nucleon, as the mass number increases the binding energy per nucleon increases, and the value for the binding energy per nucleon has a maximum value for nuclei with a mass number around 60. Verify that this is the case by determining the binding energy per nucleon for each of the following four nuclei. (Let the mass of a proton be 1.0078 u, the mass of a neutron be 1.0087 u, the mass of 6He be 6.0189 u, the mass of 8Li be 8.0225 u, the mass of 62Ni be 61.9283 u, and the mass of 115In be 114.9039 u. Enter your answers in MeV and to at least three significant figures.) (a) 6He MeV (b) 8Li MeV (c) 62Ni MeV (d) 115In MeV
Chemistry
1 answer:
Illusion [34]3 years ago
8 0

Answer:

6He =   4.90 MeV/Nucleon

8Li =     5.18  MeV / Nucleon

62Ni =   8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its  protons and neutrons, this mass is greater than  the actual  value . This is the mass defect.

Now this mass defect we can convert to energy  by utilizing Einstein´s equation, E = mc².This is  the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons =   2 x 1.0078 u  =  2.0516 u

mass neutrons = 4 x 1.0087 u  =  4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹²  J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

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2 years ago
If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?
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The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>).  If </span>Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?


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The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left, since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
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Kp is the equilibrium constant in term of the partial pressures of the gases.

Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.

Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

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3 years ago
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As we are given the mechanism for the reaction :

Step 1 : 2NO+H_2\rightarrow N_2+H_2O_2    (slow)

Step 2 : H_2O_2+H_2\rightarrow 2H_2O     (fast)

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The rate law expression for overall reaction should be in terms of NO\text{ and }H_2.

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Rate=k[NO]^2[H_2]

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8 0
3 years ago
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