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Feliz [49]
3 years ago
5

Which of the following is unchanged at the end of the CNO cycle?

Chemistry
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

The correct option is: A. carbon-12

Explanation:

The CNO cycle, the abbreviation for the carbon-nitrogen-oxygen cycle, is a catalytic cycle by which the stars produce helium from elemental hydrogen, via a series of nuclear fusion reactions.

This cycle involves the fusion of four protons with carbon (_{6}^{12}\textrm{C}), nitrogen isotope (_{7}^{13}\textrm{N}), and oxygen isotope (_{8}^{15}\textrm{O}), to give an alpha particle and two electron neutrinos and positrons.

The reaction involves the regeneration of carbon (_{6}^{12}\textrm{C}) nucleus in the last step.

_{6}^{12}\textrm{C} → _{7}^{13}\textrm{N} → _{6}^{13}\textrm{C} → _{7}^{14}\textrm{N} → _{8}^{15}\textrm{O} → _{7}^{15}\textrm{N} → _{6}^{12}\textrm{C}

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3 years ago
A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder
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Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

PV_1=nRT

where,

P = pressure of gas = 1 atm

V_1 = volume of gas = ?

T = temperature of gas = 270^oC=273+270=543K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K

V_1=9.12L

Now we have to calculate the volume at 24°C.

PV_2=nRT

where,

P = pressure of gas = 1 atm

V_2 = volume of gas = ?

T = temperature of gas = 24^oC=273+24=297K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K

V_2=4.99L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 9.12 L

V_2 = final volume = 4.99 L

Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (4.99-9.12)L

w=4.13L.artm=4.13\times 101.3J=418.4J

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

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3 years ago
A 1.00 cm2 blackbody surface emits 201 watts of radiant power. What is the temperature of the surface? Give to the nearest whole
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Explanation:

Formula for black body radiation is as follows.

                 \frac{P}{A} = \sigma \times T^{4} J/m^{2}s

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