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kykrilka [37]
2 years ago
9

A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m

icroliters must be administered?
A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many microliters must be administered?
Chemistry
1 answer:
oksian1 [2.3K]2 years ago
4 0

Answer : The 866.66\mu L must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of 3.0mCi/mL.

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in \frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L

Conversion :

(1ml=1000\mu L)

Therefore, the 866.66\mu L must be administered.

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Base pairs of DNA are A with T, and C with G

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Read 2 more answers
What amount of carbon dioxide (in moles) is produced from the reaction of 2.24 moles of ethanol with excess oxygen?
algol13

Answer : The amount of carbon dioxide produced is, 197.12 grams.

Explanation : Given,

Moles of ethanol = 2.24 mole

Molar mass of carbon dioxide = 44 g/mole

The balanced chemical reaction will be,

C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O

First we have to calculate the moles carbon dioxide.

From the balanced chemical reaction, we conclude that

As, 1 mole of ethanol react to give 2 moles of carbon dioxide

So, 2.24 mole of ethanol react to give 2\times 2.24=4.48 moles of carbon dioxide

Now we have to calculate the mass of carbon dioxide.

\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2

\text{Mass of }CO_2=4.48mole\times 44g/mole=197.12g

Therefore, the amount of carbon dioxide produced is 197.12 grams.

8 0
2 years ago
When are atoms considered to be stable?
Sergio [31]

Answer:

Atoms are at their most stable when their outermost energy level is either empty of electrons or filled with electrons.

8 0
2 years ago
A compound has a molar mass of 100 g/mol and the percent composition (by mass) of 65.45% C, 5.45% H, and 29.09% 0. Determine the
coldgirl [10]

Answer:

B) C3H3O and C6H6O2

Explanation:

Given data:

Molar mass of compound = 100 g/mol

Percentage of hydrogen = 5.45%

Percentage of carbon = 65.45%

Percentage of oxygen = 29.09%

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 5.45 / 1.01 = 5.4

Number of gram atoms of O = 29.09/ 16 = 1.8

Number of gram atoms of C = 65.45 / 12 = 5.5

Atomic ratio:

            C                      :      H            :         O

           5.5/1.8              :     5.4/1.8     :        1.8/1.8

            3                      :        3          :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 12×3 + 1.01 ×3 + 16×1 = 55.03  

n = 100 / 5503

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₃O)

Molecular formula = C₆H₆O₂

3 0
2 years ago
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