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3 years ago

How to bound Lewis dot stucture

1 answer:

6 0

1.Draw the Lewis structure.
2.Count the total number of bonds that you have.
3.Count the number of bond groups between individual atoms.
4.Divide the number of bonds between atoms by the total number of bond groups in the molecule.

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A series of chemicals were added to some AgNO3(aq). NaCl(aq) was added first to the silver nitrate solution to produce a precipi

OLEGan [10]

Answer:

First, precipitate of AgCl is formed. Second, a soluble complex of silver and ammonia is formed. Third, AgCl is reproduced due to disappearance of ammonia complex in presence of $HNO_{3}$ .

Explanation:

In presence of NaCl, $AgNO_{3}$ forms an insoluble precipitate of AgCl.

Reaction: $Ag^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)$

In presence of $NH_{3}$ , AgCl gets dissolved into solution due to formation of soluble $[Ag(NH_{3})_{2}]^{+}$ complex.

Reaction: $AgCl(s)+2NH_{3}(aq.)\rightarrow [Ag(NH_{3})_{2}]^{+}(aq.)+ Cl^{-}(aq.)$

In presence of $HNO_{3}$ , $[Ag(NH_{3})_{2}]^{+}$ complex gets destroyed and free $Cl^{-}$ again reacts with free $Ag^{+}$ to produce insoluble AgCl

Reaction: $[Ag(NH_{3})_{2}]^{+}(aq.)+2H^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)+2NH_{4}^{+}(aq.)$

4 0

3 years ago

An aqueous solution contains 0.29 M of benzoic acid (HA) and 0.16 M of sodium benzoate (A-). If the pH of this solution was meas

Inga [223]

Answer:

pKa = 4.89.

Explanation:

We can solve this problem by using the Henderson-Hasselbach equation, which states:

pH = pKa + log $\frac{[A^-]}{[HA]}$

In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

We input the given data:

4.63 = pKa + log $\frac{0.16}{0.29}$

And solve for pKa:

pKa = 4.89

3 0

3 years ago

If an equal mass of sugar and aluminum were poured into separate containers, which would require the smaller container?

kirill115 [55]

Aluminium requires the smaller container !

8 0

3 years ago

Read 2 more answers

Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of

natima [27]

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several energy values for each interaction present in the Newman structures:

-) Methyl-methyl gauche: 3.8 KJ/mol

-) Methyl-H eclipse: 6.0 KJ/mol

-) Methyl-methyl eclipse: 11.0 KJ/mol

-) H-H eclipse: 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

Molecule A

In this molecule, we have 2 Methyl-methyl gauche interactions only, so:

(3.8x2) = 7.6 KJ/mol

Molecule B

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

Molecule C

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule D

In this molecule, we have three Methyl-H eclipse interaction, so:

(6*3) = 18 KJ/mol

Molecule E

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule F

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

3 0

4 years ago

Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production

Lynna [10]

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ

Thus ΔH° of the reaction is -747.54kJ

3 0

3 years ago