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Dmitriy789 [7]
3 years ago
9

What determines a molecules VSEPR shape?

Chemistry
1 answer:
frosja888 [35]3 years ago
3 0
Using the VSEPR theory, the electron bond pairs and lone pairs on the center atom will help us predict the shape of a molecule. The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction.
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What process takes place when hydrogen-3 and hydrogen-2 combine to form
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A. Fusion

Explanation:

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Using stoichiometry, determine the mass of powdered drink mix needed to make 1.0 M solution of 100 mL
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How to do dots and cross diagrams
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Draw circles to represent the electron shell of each atom overlapping the circles where the atoms are bonded. Add dots to represent the outer electrons of one type of atom (H). Add crosses to represent the outer electrons of the other type of atom (Cl). Make sure the electrons are always in pairs.

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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
3 years ago
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