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lora16 [44]
3 years ago
7

33p is a radioactive isotope of phosphorus. can 33p be used to make a radioactive nucleotide?

Chemistry
2 answers:
Lunna [17]3 years ago
8 0
Yes, because it can replace the phosphorus atom in the phosphate group. Apex.
Agata [3.3K]3 years ago
5 0

Answer:

Yes, we can use 33p used in making radioactive nucleotide because phosphorus changes into phosphate group.

Explanation:

Phosphorus is the artificial radioactive element, mainly isotopes of Uranium and Polonium are used in making radioactive elements. 33p(Half life of 25.34 days) and 32p(Half life of 14.263 days) are the radioactive type of phosphorus.

Therefore, we can use 33p used in making radioactive nucleotide because phosphorus changes into phosphate group.            

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Which of the following describes the physical state of the layers of the Earth from the outside to the center?
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Read 2 more answers
A chemical reaction yields 3 moles of lithium hydroxide (LiOH). How many grams of
grin007 [14]

72g

Explanation:

Given parameters:

Number of moles of LiOH = 3moles

Unknown:

Mass of LiOH = ?

Solution:

A mole of  substance is a unit used to make quantitative measures in chemistry.

It is the amount of substance that contains the avogadro's number of particles.

The mole is related to mass using the expression below;

    Mass of a substance = number of moles x molar mass

Molar mass of LiOH = 7 + 16 + 1 = 24g/mol

 Mass of LiOH = 3 x 24 = 72g

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

8 0
3 years ago
URGENT!!-- Please help!
blondinia [14]

Moles of gas = 0.369

<h3>Further explanation</h3>

Given

P = 2 atm

V = 5.3 L

T = 350 L

Required

moles of gas

Solution

Ideal gas Law

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{2\times 5.3}{0.082\times 350}\\\\n=0.369

Avogadro's law : at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles  

moles of O₂ = 45% x 0.369 = 0.166

moles of Ar = 12% x 0.369 = 0.044

moles of N = 43% x 0.369 = 0.159

3 0
3 years ago
Using the 5M NaCI and 10% glucose stock solutions you generated, how would you prepare 100 ml of a solution that is both 150mM N
fredd [130]

Answer:

Take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

Explanation:

  • In order to calculate the required volume of the 5 M NaCl solution, we calculated the moles contained in a 100 mL solution that has a concentration of 150 mM:

0.1 L * 0.150 M = 0.015 moles of NaCl

With those moles we can calculated the required volume, using the concentration of the stock solution:

0.015 mol / 5 M = 0.003 L = 3 mL.

  • To make a solution that has a 1 % concentration of glucose, from a 10 % glucose solution, is the same as to make it ten times less concentrated. Thus, with a final volume of 100 mL, you would need to take 10 mL of the 10% glucose solution, because 100mL * 10/100 = 10.

So in order to prepare the solution, you would need to take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

4 0
3 years ago
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