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charle [14.2K]
3 years ago
10

A vial containing radioactive selenium-75 has an activity of 2.3 mci/ml. If 2.6 mci are required for a leukemia test, how many m

icroliters must be administered?
Chemistry
1 answer:
nika2105 [10]3 years ago
3 0

Answer is: 1130 microliters must be administered.

activity(selenium-75) = 2.3 mCi/mL.

radioactivity(Se-75) = 2.6 mCi; mCi is millicurie.

The curie (symbol: Ci) is a non-SI unit of radioactivity.

V(Se-75) = radioactivity(Se-75) ÷ activity(Se-75).

V(Se-75) = 2.6 mCi ÷ 2.3 mCi/mL.

V(Se-75) = 1.13 mL; volume.

V(Se-75) = 1.13 mL · 1000µL/mL.

V(Se-75) = 1130 µL.

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50 pennies or other coins (all coins must be the same value)
Neko [114]

Hey!

Hope this helps...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Question 1: <em>Turn 25 coins tails up. You now have 50% parent material and 50% daughter material. What is the half life of this substance? (20 Points)</em>


Answer:

Based upon the given information, we can assume that the half life of this coin material is 5 years.



Question 2: <em>Imagine that another half life has gone by. Remember that ONLY the parent material can change to the daughter material. Turn over your pennies to represent the passing of another half life. How much total parent material and how much total daughter material do you now have? You may express your answer in number of pennies or in %. (Note: Since you cannot turn over half a penny, round the actual number of pennies you would turn over up.) (20 points)</em>


Answer:

Based upon the given information from Question 2 and Question 3, we can assume that 13 additional coins would be turned over.  We started with 50, than flipped 50% of them over, and now another 50% over.  We have 12 coins left.



Question 3: <em>How many half lives will pass before you have only 3 pennies that are parent material and 47 pennies that are daughter material? (20 Points)</em>


Answer:

Based upon the given information from all the questions, and based upon our answers we can assume this:

1st half-life: 50 turned to 25 coins

2nd half-life: 25 turned to 12 coins

3rd half-life: 12 turns to 6 coins

4th half-life: 6 turns to 3 coins

So, after 4 half-lives (or after 20 years) we would have 3 coins left that have parent material.

------------------------------------------------------

<em>For future reference please do not post as complex questions like this, if you do (post a question that is similar to this) please state the following information:</em>

<em>"All questions posted in this post relate to one-another and cannot be answered without the other questions..."</em>

4 0
3 years ago
Draw the correct structure(s) for (2R,3S)‑2,3‑dibromobutane. Show stereochemistry clearly. To ensure proper grading, explicitly
velikii [3]

Answer:

b. cannot exist in optically active form

Explanation:

Stereochemistry is a branch of chemistry that involves the spatial arrangement of the atoms of molecules and studies how this affects the physical and chemical properties of such species.

The correct structure for (2R,3S)‑2,3‑dibromobutane can be seen in the image attached below. Since the compound is a meso compound due to the plane of symmetry. Thus, the compound is achiral. i.e. Compounds that are superimposable on its mirror image. The plane of symmetry is vertical inclined at 90°(i.e. perpendicular) to the page thus goes via the middle of the molecule.

6 0
3 years ago
What are the different sciences at GCSE
ivann1987 [24]

Answer:

The three sciences you would take at GCSE is Chemistry, Biology and Physics

6 0
3 years ago
Read 2 more answers
What is the percent of carbon in carbonate?
Sedaia [141]

Answer:

The percentage of carbon in a 55 % pure sample of carbon carbonate is. (Ca=40, C=12, O=16)

Explanation:

Hope this helps!!!!!!!!!!

3 0
3 years ago
How many grams of lead ii sulfide is produced when 25.0 g lead ii acetate reacts with excess hydrogen sulfide
ddd [48]

Answer: 18.42 grams of lead (II) sulfide will be produced in the given reaction:

Explanation: The reaction of lead (II) acetate and hydrogen sulfide follows:

(CH_3COO)_2Pb+H_2S\rightarrow PbS+2CH_3COOH

To calculate the moles, we use the formula:

Moles=\frac{\text{Given mass}}{\text{Molar mass}}    ....(1)

Molar mass of lead (II) acetate = 325.29 g/mol

Given mass of lead (II) acetate = 25 g

Putting values in above equation, we get:

Moles=\frac{25g}{325.29g/mol}=0.0768moles

We are given that hydrogen sulfide is present in excess, so limiting reagent is lead (II) acetate because it limits the formation of product.

By stoichiometry of the reaction,

1 moles of lead (II) acetate produces 1 mole of lead (II) sulfide

So, 0.0768 moles of lead (II) acetate will produce = \frac{1}{1}\times 0.0768 = 0.0768 moles of lead (II) sulfide.

Now, to calculate the mass of lead (II) sulfide, we use equation 1, we get:

Molar mass of lead (II) sulfide = 239.3 g/mol

0.0768mol=\frac{\text{Given mass}}{239.3g/mol}

\text{Mass of lead (II) sulfide}=18.42g

8 0
3 years ago
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