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WITCHER [35]
3 years ago
13

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>

<u>The code used is</u><u> 32405</u>

<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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Answer:

Mark has a speed of 6 MPS (Miles Per Hour)

Explanation:

It took mark 2 hours to ride his bike to his grandma's house, which was 12 miles away. I divided 2 by 12 and got 6. Remember this: mph = miles away ÷ time.

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The discovery of the electron as a subatomic particle was a result of
swat32

Answer:

4)experiments with cathode ray tubes

Explanation:

when sufficiently high voltage is applied across the electrods, current starts flowing through a stream of particles moving in the tube the negative electrode (cathode) to the positive electrode (anode). These were called Cathode Rays or Cathode Ray Particles

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3 years ago
The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment
astraxan [27]

Answer:

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

Explanation:

The gas phase reaction is as follows.

A \rightarrow B+C

The rate law of the reaction is as follows.

-r_{A}=kC_{A}

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = 10dm^{3}

Temperature "T" = 300 K

Volumetric flow rate of the reaction v_{o}=5dm^{3}s

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

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.

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

\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 is1.2s^{-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)

The feed has 50% A and 50%  inerts.

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

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

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

=0.5(2-1)=0.5

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}

The activation energy of the reaction can be calculated by using formula

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

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

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

Substitute the all values.

=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}

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

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

8 0
3 years ago
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Answer:

The proton remains the same.

Explanation:

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Oxidation is strictly on the transfer of electron(s) and not proton.

A metal that undergoes oxidation still has its protons intact otherwise it will not be called the ion of the metal since atomic number is called the proton number.

Sodium (Na) undergoes oxidation as follow:

Na —> Na+ + e-

Na is called sodium metal.

Na+ is called sodium ion.

Na has 11 electrons and 11 protons

Na+ has 10 electrons and 11 protons

From the above illustration, we can see that the protons of Na and Na+ are the same why their electrons differ because Na+ indicates that 1 electron has been loss or transferred.

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3 years ago
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