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3 years ago

If the mass of a radioactive substance is 8 grams and it has a half-life of 4 hours, how much mass remains after 8 hours?

Chemistry

2 answers:

lisov135 [29]3 years ago

6 0

Answer: 2 grams will be remaining.

Explanation: Half life is the time in which the amount of radioactive substance remains half.

For example if half life of a radioactive substance is 3 hours and we have 16 grams of it then in two hours, 8 grams will be remaining. In next two hours, 4 grams will be remaining. Similarly, in next two hours, 2 grams and further in next two hours 1 gram of the substance will be remaining.

For the given problem, the half life of the radioactive substance is 4 hours. It means the initial amount of this substance will remain half in 4 hours.

Originally we have 8 grams of the radioactive substance. In first 4 hours, 4 grams of the substance will be remaining. Now, in next 4 hours that is in total 8 hours, 2 grams of the substance will be remaining.

These problems could also be solved mathematically using the formula:

$\frac{N}{N_0}=(0.5)^n$

where, $N_0$ is the initial amount of the substance, N is remaining amount and n is the number of half lives.

number of half lives(n) = $\frac{time}{half life}$

From given information, time is 8 hours and half life is 4 hours.

So, n = $\frac{8}{4}$ = 2

Initial amount is given as 8 grams and it asks to calculate the remaining amount. Let's plug in the values in the equation:

$\frac{N}{8}=(0.5)^2$

$\frac{N}{8}$ = 0.25

N = 8(0.25)

N = 2

So, from both ways 2 grams of the radioactive substance will be remaining after 8 hours.

Alborosie3 years ago

3 0

The sample will lose half of its mass after 4 hours. The half life.
82=4
The sample will lose half of the remaining four after another half life.
42=2 Hope this helps! :)

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3 years ago

Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be p

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$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

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$\text{Number of moles}=\frac{5000g}{100g/mol}=50moles$

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3 0

3 years ago

Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met,

erma4kov [3.2K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

Explanation:

The acid-catalyzed hydrolysis from makes us to understand that the polypeptides contains 8 amino acids

From the question we are told that the Edman's reagent releases Leu it means that the N-terminal amino acid would be Lue(i.e Leucine)

Also from the question we are told that the Carboxypeptidase released Ser this mean the the C-terminal amino acid would be Ser ( i.e Serine)

The Amino acids would in the polypeptides would be arranged like this

Lue __ __ __ __ __ __ Ser

Now from the question we are told that treatment with cranogen bromide form two peptides.

Now generally cranogen bromide divides a peptide on the C- side(i.e the extreme left side ) of Met(Methionine)(This is an amino acid obtained by hydrolysis of most common peptides )

So this means that any peptides containing Met(Methionine) must have Methionine as a C- terminal amino acid(i.e at extreme left) and for peptides that does not contain Met must be C - terminal peptides

From the question we see that it is the second peptide that contain Met and it is a penta peptide(i.e it contains 5 amino acid)

Thus the fifth amino acid is Met

So the sequence of these amino acid would now be

Lue __ __ __ Met __ __ Ser

From the question we are told that the the Trypsin-catalyzed hydrolysis forms two amino acid and two peptides

Now generally Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine) and Lys (lysine) and any peptide that holds Arg or Lys must have them as their C- terminal amino acids

From the first peptide in the two peptide formed we see that Arg would be the Seventh amino acid of the octapeptide because commonly the trypsin that sticks to the C-side of Arg would for Ser

and Phe would be the sixth amino acid of the octapeptide

So the sequence of these amino acid would be

Lue __ __ __ Met Phe Arg Ser

Looking at the first amino acid formed from the Trypsin-catalyzed hydrolysis we see that Arg would be the fourth amino acid of octapeptide as Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine).

From the second peptide of the Trypsin-catalyzed hydrolysis we see that Lys would be the third amino acid of the octapeptide as trypsin divides on the C- side of Lys (lysine) and Tyr would be the second amino acid of the octapeptide

So the sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

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3 years ago