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Llana [10]
3 years ago
5

The isotope of an atom containing 31 protons and 39 neutrons suddenly has two neutrons added to it. what isotope is created? loo

k up the peridoic table to find which element has 31 protons.
Gallium-72
yttrium-41
yttrium-72
Gallium-41
Chemistry
2 answers:
Fed [463]3 years ago
5 0
It is gallium look it up its what google says
Rashid [163]3 years ago
3 0

Answer: Gallium-72

Explanation:

Atomic number is defined as the number of protons or number of electrons that are present in an atom. It is characteristic of an element.

Atomic number = Number of electrons = Number of protons = 31

Thus the element with atomic number of 31 is Gallium.

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

There were 39 neutrons and then 2 more were added , thus there are 39+2 = 41 neutrons

Thus mass number = 31 + 41 = 72

Thus the element will be gallium with 72 mass number.

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<u>Answer:</u> The given solution is unsaturated solution

<u>Explanation:</u>

We are given:

Solubility of NaBr = 9.19 m

Unsaturated solution is defined as the solution in which more solute particles can be dissolved in the solvent.

Saturated solution is defined as the solution in which no more solute particles can be dissolved in the solvent.

A precipitate is defined as the insoluble salt which is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

There are three conditions:

  • When m_{\text{(calculated)}}; the solution is unsaturated
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To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}  

Where,

m_{solute} = Given mass of solute (sodium bromide) = 91.3 g

M_{solute} = Molar mass of solute (sodium bromide) = 103 g/mol

W_{solvent} = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

\text{Calculated molality of NaBr}=\frac{91.3\times 1000}{103\times 115}\\\\\text{Calculated molality of NaBr}=7.71m

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