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ipn [44]
3 years ago
15

Why is it important to have the correct bond angles of the different atoms and the shape of the molecule? (20 points NEED ANSWER

ASAP)
Chemistry
1 answer:
hjlf3 years ago
5 0

It is important to have the correct bond angles of the different atoms and the shape of the molecule due to following reasons;

                  Among other properties the polarity of compounds mainly depend upon the shape and bond angles of that particular compound. For example, considering the molecule of water, we already know that it is a polar molecule with partially positive hydrogen atoms and partially negative oxygen atoms and acts as universal solvent. The bond angle in water is about 104.5° with a Bent geometry. Unlike carbon dioxide (CO₂) which has Linear structure with bond angle 180° and is non-polar in nature therefore, the bent geometry in water is responsible for the polarity.

                  Other properties which can also be predicted by predicting the bond angles along with molecular geometries are;

                                      i) Magnetism

                                      ii) Phase of matter

                                       iii) Color

                                      iv) Reactivity

                                       v) Biological activities <em>e.t.c</em>

                     

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What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)
Pie

Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

Where

pH = acidity of a buffer solution

pKa = negative logarithm of Ka

Ka =acid disassociation constant

[HA]= concentration of an acid

[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414

8 0
1 year ago
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

brainly.com/question/5976752

brainly.com/question/13707449

brainly.com/question/13707449

brainly.com/question/6263007

brainly.com/question/14641878

brainly.com/question/2912965

#SPJ1

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What is the the term for the final products in a reaction ?
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Answer:

the final product is called a product

Explanation:

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A solution is prepared from 8 grams of acetic acid ( CH3COOH ) and 725 grams of water.What is the molality of this solution?0.00
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Answer:

its 725 grams 0.0002 725.000

Explanation:

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