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2 years ago

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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True or false. adding more neutrons to a neutrally charged atom will not change the charge

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The statement is true.

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3 years ago

A teacher wrote the following part of a balanced chemical equation: Cu+2AgNO 3 To show the conservation of matter in the chemica

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2 years ago

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This isn't a homework question but how do you get bruises off of your chest in like two days??

romanna [79]

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2 years ago

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:

$t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}$

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

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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.

Explanation:

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.

It could have emitted any form of radioactive particles which can be alpha or beta.
We do not know if it has a long or short half life because the value is not given.
But since the radioactive decay in one step produces a stable isotope, we can conclude that it did not participate in a decay series.
A decay series involves a radioactive decay in multiple steps.

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3 years ago

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