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-BARSIC- [3]
2 years ago
14

Which of these processes removes carbon from

Chemistry
1 answer:
GenaCL600 [577]2 years ago
3 0
Answer is c photosynthesis
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The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be
Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature T_1 = 1001 K

Final temperature T_2 = 298 K

Applying the equation of Arrhenius theory.

In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})

where ;

R gas constant = 8.314  J/K/mol

In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})

In \dfrac{k_2}{0.380 }= -70.8655

\dfrac{k_2}{0.380 }= e^{-70.8655}

\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}

{k_2}= 1.67303256 \times 10^{-31} \times {0.380 }

{k_2}= 6.3575 \times 10^{-32}  /M .sec

Half life:

At 1001 K.

t_{1/2} = \dfrac{In_2}{k_1}

t_{1/2} = \dfrac{0.693}{0.38}

t_{1/2} = 1.82368 secc

At 298 K:

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t_{1/2} =1.09 \times 10^{31} \ sec

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3 years ago
A radioisotope forms a stable isotope after it undergoes radioactive decay. Which is the best conclusion to draw from this state
Black_prince [1.1K]

Answer:

C. It does not participate in a decay series.

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From this statement, we can deduce that a radioisotope that forms a stable isotope after it undergoes radioactive decay suggests that it does not participate in a decay series.

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  • But since the radioactive decay in one step produces a stable isotope, we can conclude that it did not participate in a decay series.
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