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horsena [70]
1 year ago
7

Optimal contrast, density and penetration have been achieved on ap abdominal projections when which anatomic structures are demo

nstrated?
Chemistry
1 answer:
Alina [70]1 year ago
8 0

Psoas major muscles, kidneys, inferior ribs, and lumbar transverse processes are demonstrated on AP abdominal projections.

<h3>What are AP abdominal projections?</h3>

The anteroposterior (AP) radiograph the patient in a supine position is the basis of the majority of plain-film examinations of the abdomen.

However, an upright-position film could also be helpful in patients with suspected bowel obstruction, also as to assess the air-fluid levels in the distended bowel.

This information also can be obtained on a film of the abdomen taken with the patient in the lateral decubitus position.

An AP supine projection is usually called a flat plate or KUB (because it includes the kidneys, ureter, and bladder)

To learn more about AP projections: brainly.com/question/14903835

#SPJ4

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Explanation:

Because you are epic.

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Use the periodic table and example bond table to answer the question.
lyudmila [28]

In keeping with the general trends, K-Br will have the smallest bond energy. The bond energy refers to the energy that keeps the atoms in a bond together.

<h3>What is bond energy?</h3>

Bond energy is the energy that is required to hold atoms together in a bond. This energy must also be supplied when the atoms are to be separated.

We have the bond energies of each of the bons in the question, we have to note that the the smallest value of bond energy is Na-Br hence in keeping with the general trends, K-Br will have the smallest bond energy.

Learn more about bond energies: brainly.com/question/14842720?

7 0
2 years ago
4. Imagine you have a radioactive sample containing both Virtualium and decayed atoms of Virtualium. After analysis, you find it
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7 0
3 years ago
Complete combustion of 2.60 g of a hydrocarbon produced 8.46 g of co2 and 2.60 g of h2o. what is the empirical formula for the h
zubka84 [21]
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon. 
Hence, in this case the mass of carbon in 8.46 g of CO2:
 (12/44) × 8.46 = 2.3073 g
 1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
 (2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12  = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
             Carbon : Hydrogen
  0.1923/0.1923 : 0.2889/0.1923
                       1 :  1.5
                      (1 : 1.5) 2
                     = 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
5 0
4 years ago
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