The cyclo-heptatrienyl anion will be considered as antiaromatic compound.
A cyclic compound with a greater energy electron system due to the existence of a 4n delocalized electron is what makes up an anti-aromatic chemical.
Due to its eight pi electrons, the cycloheptatrienyl anion (also known as the tropylium anion) should be antiaromatic; yet, the extra lone pair on one carbon could enable that carbon to become 
hybridized, placing the extra electrons in an orbital. It would become non-planar as well as non-aromatic as a result.
Therefore, the cyclo-heptatrienyl anion will be considered as antiaromatic compound.
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Answer:
4.52 grams of potassium chloride
Explanation:
First you need a balanced reaction equation K(s) + Cl₂ (g) --> 2 KCl(s). Since the chlorine gas is said to be in excess all of the potassium will be converted to form potassium chloride. Convert grams of potassium to moles of potassium with the periodic table, then you can convert moles of potassium to moles of potassium chloride with the balanced equation, finally convert moles of potassium chloride to grams of potassium chloride with the periodic table. Set up a dimensional analysis chart to help with conversions
(2.50 grams K) * (<u>1 mole K</u>) * (<u>2 mole KCl)</u> * (<u>35.45 grams KCl</u>)
(39.09 grams K) (1 mole K) (1 mol KCl)
Which equals 4.52440778 grams, with sig figs it should be 4.52 grams of potassium chloride
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For CaCl₂,
E.N of Chlorine = 3.16
E.N of Calcium = 1.00
________
E.N Difference 2.16 (Ionic Bond)
For C-H Bond in C₆H₁₂O₆,
E.N of Carbon = 2.55
E.N of Hydrogen = 2.20
________
E.N Difference 0.35 (Non-Polar Covalent Bond)
For C-O Bond in C₆H₁₂O₆,
E.N of Oxygen = 3.44
E.N of Hydrogen = 2.55
________
E.N Difference 0.89 (Polar Covalent Bond)
For O-H Bond in C₆H₁₂O₆,
E.N of Oxygen = 3.44
E.N of Hydrogen = 2.20
________
E.N Difference 1.24 (Polar Covalent Bond)
For MgO,
E.N of Oxygen = 3.44
E.N of Magnesium = 1.31
________
E.N Difference 2.13 (Ionic Bond)
For Na₂O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For SiO₂,
E.N of Oxygen = 3.44
E.N of Silicon = 1.90
________
E.N Difference 1.54 (Polar Covalent Bond)
Result:
Compounds having Ionic Bonds are Na₂O, MgO and CaCl₂.
Answer:
3.76 x 1014 s-1? λ = c/ν = 3.00 x 108 m/s = 7.98 x 10-7 m 3.76 x 1014 s-1.
Explanation:
hope i helped btw i am breanna plzz mark me brainliest
Answer:
Kindly check the attached pictures.
Explanation:
For the reaction between ammonium salt and Na2CO3, in the reaction mechanism, there are going to be four steps for the production of nicotine. The chemical species to monitor here are the Na2CO3 and the starting material which has the amine group attached to it.
STEP ONE: Dissociation of the Na2CO3.
Na2CO3 <=========> 2Na^+ + CO3^2-
STEP TWO: (will be shown in the attached picture). It is the abstraction of H^+ from the amine in the starting material.
STEP THREE: (will be shown in the attached picture). It is the formation of a cyclic compound due to nucleophilic reaction.
STEP FOUR: (will be shown in the attached picture). It involve the the deprotonation and this is the final phase.
