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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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How many unpaired electrons are present in a ground-state atom from each of the following groups?

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For 7A(17) :

Electronic configuration $ns^2 np^5$

So, there are 5 unpaired electrons present in group 7A(17).

<h3>What are Unpaired Electrons?</h3>

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2 years ago

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Answer:

Please find how these data prove the law of multiple proportions below

Explanation:

The law of multiple proportions was proposed by an English chemist called John Dalton. The law states that when two elements combine and to form more than one compound. The weights/masses of the second element in the two compounds, which combines with a fixed ratio of the first element, is in a simple whole number ratio.

In this question, Nitrogen is said to combine with oxygen to give three different compounds as follows:

A) laughing gas containing 63.65% nitrogen i.e. 0.6365g

This means that the mass of oxygen will be (1-0.6365) = 0.3635g

B) colorless gas containing 46.68% nitrogen i.e. 0.4668g

This means that the mass of oxygen will be 0.5332g

C) brown toxic gas containing 30.45% nitrogen i.e. 0.3045g

This means that the mass of oxygen will be 0.6955g

The ratios of oxygen in the three compounds is therefore:

0.3635: 0.5332: 0.6955

Divide this ratio by the smallest number (0.3635)

0.3635/0.3635 = 1

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0.6955/0.3635 = 1.913

Multiply this ratio by 2, we have:

2: 2.9 : 3.8

Hence, the simple whole number ratio is 2:3:4.

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3 years ago

Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

b) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

c) The IUPAC name is Butan-2-ol

Explanation:

Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.

According to that for molecule 1 as shown on the second uploaded image

For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

Considering the third question

The name of the given molecule is gotten according to longest carbon chain = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

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3 years ago