10.0 g of Mg are reacted with 95.0 g of I2. If 27.5 g of magnesium iodide are obtained, what is the percent yield

Answer:

(a) The nitrogen atom in contains 8 electrons surrounding the central nitrogen atom.

(b) The nitrogen atom in has 8 electrons around the central nitrogen atom.

(c) The nitrogen atom in has 8 electrons surrounding the central nitrogen atom.

Explanation:

The Lewis dot structure illustrates the connections between atoms within a molecule, additionally showing unpaired electrons present in the molecule.In this structure, valence electrons are represented as 'dots'.

Now we will determine the number of electrons linked to the central nitrogen atom in the specified molecule.

(a) The identified molecule is,

Recognizing that nitrogen has '5' valence electrons while hydrogen has '1', we ascertain the total valence electrons in

= 5 + 2(1) + 1 = 8

According to the Lewis dot structure, this reveals 4 bonding and 4 non-bonding electrons.

From the Lewis structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

(b) The identified molecule is,

Knowing that nitrogen has '5' valence electrons and hydrogen has '1' valence electron, the total number of valence electrons for = 5 + 4(1) - 1 = 8

In the Lewis dot structure, there are 8 bonding electrons and 0 non-bonding electrons.

This leads us to conclude that the nitrogen atom in

contains 8 electrons surrounding it.

(c) The identified molecule is,

Recognizing nitrogen’s '5' valence electrons and hydrogen’s '1' brings the total valence electrons for

= 5 + 3(1) = 8Utilizing the Lewis dot structure indicates there are 6 bonding and 2 non-bonding electrons.

According to the Lewis dot structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

The response is:

No, the equation is not balanced. Neither the Nitrogen (N) nor the Hydrogen (H) are in balance!

Here's the reasoning:

⓵ A properly balanced chemical equation means that the quantity of atoms on the reactants side matches that on the products side.

→ The equation lacks balance because there are 2 Nitrogen atoms and 2 Hydrogen atoms on the reactants side. In contrast, on the products side, there is only 1 Nitrogen atom and 4 Hydrogen atoms. Thus, the number of atoms on each side is not consistent!

Hopefully, this clarification is helpful; feel free to reach out if you have any further questions! ☻

Answer:

                    Benzoyl Chloride

Explanation:

                     Friedel-Craft Acylation is a reaction used to incorporate electrophiles (especially carbonyl groups) into the double bonds of aromatic compounds.

This scenario involves synthesizing benzophenone using benzene and benzoyl chloride.

Mechanism:

Step 1: Initially, benzoyl chloride reacts with a Lewis acid to create an electrophile known as oxo(phenyl)methylium.

Step 2: The second stage involves benzene participating in an electrophilic substitution reaction, where the electrophile is added.

Step 3: Finally, the aromatic nature of benzene is restored by releasing a proton.

Answer:

Empirical formula: Na3S4O9

Molecular formula: Na6S8O18

Explanation:

1) Upang makuha ang empirical formula, na siyang pinaka-simpleng buong bilang na ratio ng bawat elemento sa hindi kilalang compound na ito, una nating hahatiin ang dami ng bawat elemento sa kanilang sariling atomic mass.

Na= 23g/mol, S= 32g/mol, O=16g/mol

Na = 20.2/23 = 0.878mol

S = 37.6/32 = 1.175mol

O = 42.2/16 = 2.638mol

Susundan ito sa pamamagitan ng paghahati ng bawat bilang ng moles sa pinakamababang halaga (0.878mol)

Na = 0.878/0.878 = 1

S = 1.175/0.878 = 1.34

O = 2.638/0.878 = 3.004

Susunod, paramihin ang bawat decimal na bahagi ng 3 upang maging buo ang mga numero:

Na = 3

S= 4.02 ~ 4

O= 9.012 ~ 9

Kaya, ang pinaka-simpleng ratio ng Na, S, at O ay 3:4:9 kaya, ang empirical formula ay Na3S4O9

2) Para makuha ang molecular formula, gagamitin natin ang molecular weight ng bawat elemento sa empirical ratio, at ang kabuuang molar mass ng compound i.e.

Molar mass ng compound= 682.8g/mol

molecular weight ng Na3S4O9 = (3 x 23g/mol) + (4 x 32 g/mol) + (9 x 16 g/mol)

= 69 + 128 + 144

= 341 g/mol

Kaya, (Na3S4O9)n = 682.8g/mol

= (341)n = 682.8

n = 682.8/341

n~ 2

Kaya, molecular formula = (Na3S4O9)2

= Na6S8O18

Refer to the explanation and the attached image for further details. When AlCl3 reacts with (CH3)3CCH2Cl, a primary carbocation is produced. This primary carbocation then undergoes a 1,2-alkyl shift leading to the formation of a tertiary carbocation, which subsequently bonds with the benzene ring as depicted in the attached image.