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3 years ago

What CPT code is reported for a percutaneous needle biopsy of mediastinum?

Chemistry

2 answers:

Klio2033 [76]3 years ago

7 0

<h2>Answer:</h2>

<h2>Explanation:</h2>

CPT stands for “Current Procedural Terminology”. Every medical, diagnostic, or surgical procedure or service has an associated 5-digit CPT code assigned to it. CPT code 32405 is reported for a percutaneous needle biopsy of mediastinum. The Current Procedural Terminology (CPT) code 32405 as maintained by American Medical Association, is a medical procedural code under the range - Excision/Resection Procedures on the Lungs and Pleura.

AveGali [126]3 years ago

5 0

Biopsy, lung or mediastinum, percutaneous needle

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3 years ago

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Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

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The rate law of the reaction is as follows.

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Temperature "T" = 300 K

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$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

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$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

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Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

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Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

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$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

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$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

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$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

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Answer:

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