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babunello [35]
2 years ago
14

Fluoride is moderately basic, by far the most basic of the halides. It is a terrible leaving group. Fluoride is many orders of m

agnitude better than bromide in leaving group ability. Bromide is many orders of magnitude better than fluoride in leaving group ability. Bromide ion is so weak it is not considered at all basic; it is an excellent leaving group.
Chemistry
1 answer:
TEA [102]2 years ago
3 0

Answer:   Bromide is many orders of magnitude better than fluoride in leaving group ability

Explanation:

As Size of an atom  Increases,  the Basicity Decreases this is because  if we move downwards  from the top of the periodic table to the bottom of the periodic table, the size of an atom increases. As size increases, basicity will decrease, meaning the element  will be less likely to act as a base implying that the element will be less likely to share its electrons.

in the same vein. With an increase in size, basicity decreases, making the ability of the leaving group to leave increase to increase . This can be seen in the halogens going down the group  from

F--- worst

Cl----fair

Br ----good

 I-----excellent

with fluorine having the worst ability to leave than Bromine which is better in terms of the leaving group ability.

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What is the molality of a solution in which 0.42 moles aluminum chloride has been dissolved in 4200 water ?
madreJ [45]

Answer:

0.1 M

<h3>Explanation:</h3>
  • Molarity refers to the concentration of a solution in moles per liter.
  • It is calculated by dividing the number of moles of solute by the volume of solvent;
  • Molarity = Moles of the solute ÷ Volume of the solvent

<u>In this case, we are given;</u>

  • Number of moles of the solute, NH₄Cl as 0.42 moles
  • Volume of the solvent, water as 4200 mL or 4.2 L

Therefore;

Molarity = 0.42 moles ÷ 4.2 L

            = 0.1 mol/L or 0.1 M

Thus, the molarity of the solution will be 0.1 M

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3 years ago
I need help with unit conversion​
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Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th
stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have:  2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH=2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}

Knowing:

  • ΔH= 75.5 kJ/mol
  • H_{NO}= 90.25 kJ/mol
  • H_{Cl_{2} }= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound  the chlorine Cl₂)
  • H_{NOCl}=?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - H_{NOCl}

Solving

-H_{NOCl}=75.5 kJ/mol - 2*90.25 kJ/mol

-H_{NOCl}=-105 kJ/mol

H_{NOCl}=105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0
2 years ago
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