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Juli2301 [7.4K]
3 years ago
11

Halogenated compounds are particularly easy to identify by their mass spectra because chlorine and bromine occur naturally as mi

xtures of two abundant isotopes. Chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%); bromine occurs as 79Br (50.7%) and 81Br (49.3%). For the compound Chlorocyclohexane, C6H11Cl: At what masses do the molecular ions occur? (List in order of increasing mass separated by commas, e.g. 120,122.) What are the percentages of each molecular ion?
Chemistry
1 answer:
12345 [234]3 years ago
5 0

Answer:

The increasing order of masses of molecule ions:

118 g/mol(75.8%) ,  120 g/mol(24.2%)

Explanation:

Chlorine occurs as 35-Cl (75.8%) and 37-Cl (24.2%).

Atomic mass of 35-Cl = 35 g/mol

Atomic mass of 37-Cl = 37 g/mol

Mass of Chlorocyclohexane in which 35-cl is present as a chlorine atom: C_6H_{11}Cl

=6\times 12 g/mol+11\times 1 g/mol+1\times 35 g/mol=118 g/mol

Mass of Chlorocyclohexane in which 37-Cl is present as a chlorine atom: C_6H_{11}Cl

=6\times 12 g/mol+11\times 1 g/mol+1\times 37 g/mol=120 g/mol

The increasing order of masses of molecule ions:

118 g/mol(75.8%) < 120 g/mol(24.2%)

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A 3.25 L solution is prepared by dissolving 285 g of BaBr2 in water. Determine the molarity.
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Answer:

0.295 mol/L

Explanation:

Given data:

Volume of solution = 3.25 L

Mass of BaBr₂ = 285 g

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of solute:

Number of moles = mass/ molar mass

Molar mass of BaBr₂ = 297.1 g/mol

Number of moles = 285 g/ 297.1 g/mol

Number of moles= 0.959 mol

Molarity:

M = 0.959 mol / 3.25 L

M = 0.295 mol/L

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The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?
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Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

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  • O: 16 g/mole

the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

moles of water=\frac{3.11 grams*1 mole}{18 gramos}

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

heat=\frac{0.1728 moles*40.66 kJ}{1 mole}

heat= 7.026 kJ

<u><em>The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.</em></u>

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