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Marina CMI [18]
2 years ago
7

Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Chemistry
1 answer:
Virty [35]2 years ago
5 0

Thee question is incomplete; the complete question is;

Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Select one or more:

A: Chlorine

B: Carbon

C: Hydrogen

D: Sulfur

E: Fluorine

F: Oxygen

Answer:

chlorine

sulphur

Explanation:

The octet rule states that, for  atoms to be stable, they must have eight electrons on their outermost shells.

This rule is not strictly followed  by some elements such as sulphur and chlorine. The atoms of these elements can sometimes expand their octet by utilizing the d-orbitals found in the third principal energy level and beyond.

These leads to formation of compounds in which the central atom has more than eight electrons in its outermost shell.

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The percent composition by mass of a compound is 76.0% c, 12.8% h, and 11.2% o. the molar mass of this compound is 284.5 g/mol.
Leokris [45]
You have a few steps to solve this one. First, we'll find the molar mass by percentage of each element in the molecule. Then, we'll divide each of those relative masses by the atomic mass of each element. The number of times the mass divides into the relative mass is the number of atoms of that element in the molecule:

C: 284.5 x .76 = 216.22
H: 284.5 x .128= 36.416
O: 284.5 x .112 = 31.864.

Now we divide out each element's atomic mass (from the periodic table). it's okay if they're approximated from the decimal answer.
C: 216.22 ÷ 12.011 ≈ 18
H: 36.416 ÷ 1.008 ≈36
O: 31.864 ÷ 15.999 ≈ 2

Therefore, the molecular formula is C18H36O2. 

The empirical formula would be found by dividing out all factors of those subscript numbers. In our case, all of them can be divided by 2. The empirical formula would be C9H18O




7 0
2 years ago
Lightsticks produce light via a chemical reaction. Dropping a lightstick into hot water makes it glow A) less intensely. B) more
victus00 [196]

Answer:

B.) More intensely

Explanation:

If you immerse a lightstick in hot water, the chemical reaction will speed up. The stick will glow much more brightly but will wear out faster too.

4 0
2 years ago
Read 2 more answers
A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoi- chiometric point.
Dmitrij [34]

Answer:

Molar concentration of the weak acid solution is 0.0932

Explanation:

Using the formula: \frac{C_aV_a}{C_bV+b}  = \frac{n_a}{n_b}

Where Ca = molarity of acid

Cb = molarity of base = 0.0981 M

Va = volume of acid = 25.0 mL

Vb = volume of base = 23.74 mL

na = mole of acid

nb = mole of base

Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1

Therefore, C_a = \frac{C_bV_b}{V_a}

Ca = 0.0981 x 23.74/25.0

                 = 0.093155 M

To 4 significant figure = 0.0932 M

3 0
2 years ago
Read 2 more answers
What is the oxidation number of pt in k2ptcl6?
padilas [110]
Your compound is K_{2}PtCl_{6}.

Remember that the oxidation numbers in a neutral compound must add up to zero. Cl has an oxidation number of -1 because it is a halogen K has an oxidation number of +1 because it is an alkali metal, which exhibits an oxidation state of +1 in compounds.

Since you have 6 atoms of Cl, you have -1(6) = -6 for the Cl. Since you 2 atoms of K, you have +1(2) = +2 for the K. The oxidation number of Pt must make all the oxidation numbers add up to zero:
+2 + (-6) + oxidation number of Pt = 0
-4 + oxidation number of Pt = 0
Oxidation number of Pt = 4
4 0
2 years ago
Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.
Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0
2 years ago
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