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Rzqust [24]
3 years ago
13

What radionuclides is used in diagnosis of atherosclerosis

Chemistry
1 answer:
aliina [53]3 years ago
8 0
Answer: technetium-99.

Explanation:

This is not an answer that you can look for in a book.

You need to do some research.

You can fiind the abstract of the article titled <span>Molecular imaging of atherosclerosis using a technetium-99m-labeled endothelin derivative.</span>

The conclusion of this scientific article is that labeling with the radiosotope Tc - 99 iis feasibele to visualize the effects of an experiment inducing atherosclerosis, which gives the answer to the question posted.
.

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How many elements are in the following equation? salt pure water(NaC/<br> + H2O)
AlexFokin [52]

Answer:

4

Explanation:

Sodium Na, Carbon C, hydrogen H2, Oxygen O

6 0
3 years ago
Read 2 more answers
What is the difference between the Lewis model and the valence-shell electron pair repulsion (VSEPR) model?
grandymaker [24]

Answer: only the VSEPR mode shows the geometric shape of a formula.

8 0
3 years ago
Assuming a car (with a 70-L) gas tank can hold approximately 50,000 (5.00 * 10^4) g of octane(C8H18) or 50,000 (5.00 * 10^4) g o
konstantin123 [22]

Answer:

- From octane: m_{CO_2}=1.54x10^5gCO_2

- From ethanol: m_{CO_2}=9.57x10^4gCO_2

Explanation:

Hello,

At first, for the combustion of octane, the following chemical reaction is carried out:

C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O

Thus, the produced mass of carbon dioxide is:

m_{CO_2}=5.00x10^4gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{8molCO_2}{1molC_8H_{18}}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=1.54x10^5gCO_2

Now, for ethanol:

C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O

m_{CO_2}=5.00x10^4gC_2H_6O*\frac{1molC_2H_6O}{46gC_2H_6O}*\frac{2molCO_2}{1molC_2H_6O}*\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=9.57x10^4gCO_2

Best regards.

3 0
3 years ago
Is this a scientific model? Use complete sentences to explain why or why not
Fittoniya [83]

Answer:

it is

Explanation:

8 0
3 years ago
The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo
noname [10]

Answer:

<h2>36.09 L</h2>

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume.

Since we're finding the initial volume

V_1 =  \frac{P_2V_2}{P_1}  \\

We have

V_1 =  \frac{7.98 \times 14.2}{3.14} =   \frac{113.316}{3.14}  \\  = 36.0878...

We have the final answer as

<h3>36.09 L</h3>

Hope this helps you

8 0
2 years ago
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