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Rudiy27
3 years ago
9

A 50 gram sample of a radioisotope undergoes 2 half-lives. How many grams

Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

At the end of second half life 12.5 g will left

Explanation:

Given data:

Total Mass = 50 g

Half lives = 2

Mass remain at the end = ?

Solution:

At time zero = 50 g

At 1st half life = 50 g /2 = 25 g

At second half life = 25 g/2 = 12.5 g

So at the end of second half life 12.5 g will left.

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Write the formula for methionine (red = o, gray = c, blue = n, yellow = s, ivory = h).express your answer as a molecular formula
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We have to get the formula of methionine.

The formula of methionine is

Methionine is an essential amino acid of human. It is white crystalline solid.

The formula of methionine is C₅H₁₁NO₂S. Methionine contains five carbon atoms, eleven hydrogen atoms, one nitrogen atom, two oxygen atoms and one sulphur atom.

It contains one amino group (-NH₂) and one carboxylic acid group (-COOH). Atoms are depicted in C-atom in grey, H-atom in ivory, O-atom in red, N-atom in blue and S-atom in yellow.

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3 years ago
Can someone help me with this please
never [62]

Answer: B. 1:2

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3 0
3 years ago
Describe the valence electrons in nitrogen and how it could bond to other atoms.
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Answer:

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Explanation:

Sorry i had to look it up i didn't know this answer

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6 0
3 years ago
A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state.Th
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3 years ago
Determine the theoretical maximum moles of hydroquinone, , that could be produced in this experiment. The reactant, quinone, is
dedylja [7]

Answer:

Theoretical maximum moles of hydroquinone: 0.2167 mol.

Explanation:

Hello,

In this case, the undergoing chemical reaction is like:

Quinone\rightarrow Hydroquinone

In such a way, since the molar mass of quinone is 108.1 g/mol and it is in a 1:1 molar ratio with hydroquinone, we can easily compute the theoretical maximum moles of hydroquinone by stoichiometry:

n_{hydroquinone}=23.4g\ quinone*\frac{1mol\ quinone}{108.1 g\ quinone}*\frac{1mol\ hydroquinone}{1mol\ quinone}  \\\\n_{hydroquinone}=0.2167mol

Clearly, this is the theoretical yield which in grams is:

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Y=\frac{13.0g}{23.84g}*100\%=54.5\%

Best regards.

8 0
3 years ago
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