Why do health care professionals limit the number of radiation treatments that a person can be given

1 answer:

6 0

Radiation is s very powerful substance and can cause major health problems do the health care professionals limit the use to help not create sickness

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Can a cold front warm air moves which direction

Katena32 [7]

East is the cold front

6 0

3 years ago

when two amino acid monomers are positioned so that the carboxyl group of one is adjacent to the amino group of the other, they

zalisa [80]

They can be joined through a Peptide bond

The amino acids serve as the foundation for proteins. Water is produced when the amino acids are linked to form a lengthy chain of acids via amino and carboxyl. The main protein is made up of these long chain amino acids.

When the carboxyl group of one molecule combines with the amino group of the other molecule, a molecule of water is released, and a peptide bond is created between the two molecules (H2O). This condensation event, sometimes referred to as a dehydration synthesis reaction, typically takes place between amino acids.

<h3>What is a Peptide bond ?</h3>

The carboxyl group of one amino acid is joined to the amino group of another to produce a peptide bond, also known as a eupeptide bond. In essence, a peptide link is an amide-type covalent chemical bond.

Learn more about Peptide bond here:

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8 months ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$