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2 years ago

What type of orbitals overlap to provide stability to the tert-butyl carbocation by hyperconjugation

Chemistry

1 answer:

larisa86 [58]2 years ago

7 0

Explanation:

The - 3 degree C( carbon atom) 2p atomic orbital + methyl C-H sigma molecular orbital because one C-H bond has to dissolve its bond and provide the H that is sigma molecular orbital and the carbonation is type 3 degree sp2 carbon.

Hyperconjugation is the stabilizing effect arising from the electrons ' engagement in a π-bond (usually C-H or C-C) with a neighboring empty or partly filled p-orbital or π-orbital to provide an expanded molecular orbital that enhances system stability.

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What is the axmen classification for benzene (C6H6)?

Tatiana [17]

Answer is: Benzene is trigonal (or triangular) planar.

VSEPR theory (The Valence Shell Electron Pair Repulsion Theory) uses the AXE notation (m and n are integers, m + n = number of regions of electron density).

For benzene molecule (C₆H₆):

m = 3; the number of atoms bonded to the central atom.

n = 0; the number of lone pairs on the central atom.

8 0

1 year ago

Read 2 more answers

An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the or

galben [10]

The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s

4 0

2 years ago

Read 2 more answers

Consider the reaction: 2 Al + 3Br2 → 2 AlBr3 Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br2. What is

timama [110]

Answer:

4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel

Explanation:

The reaction given in the question is

$2Al + 3 Br_2$ ⇒ $2AlBr_3$

According to the stoichiometric coefficients of the reaction, 2 moles of $Al$ requires 3 moles of $Br_2$ so in this reaction, $Br_2$ is a limiting reagent. So we will consider that $Al$ is in excess.

Now,

Since 3 moles of $Br_2$ requires 2 moles of $Al$

So, for 6 moles of $Br_2$ the moles of $Al$ required = $\frac{2}{3} \times 6$ = 4 moles.

Moles of $Al$ remaining after the completion of reaction = 5 - 4 = 1 mole.

Again,

Since 3 moles of $Br_2$ produces 2 moles of $AlBr_3$

So, moles of $AlBr_3$ produced by 6 moles of $Br_2$ = $\frac{2}{3} \times 6$ = 4 moles.

Therefore, after the completion of reaction, 4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel.

5 0

2 years ago

What mass of boron sulfide must be processed with 2.1 x 10 4g of carbon to yield 3.11 x 10 4 g of boron and 1.47 x 10 5 g of car

Basile [38]

The reaction between boron sulfide and carbon is given as:

2B2S3 + 3C → 4B + 3CS2

As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.

Given data:

Mass of C = 2.1 * 10^ 4 g

Mass of B = 3.11*10^4 g

Mass of CS2 = 1.47*10^5

Mass of B2S3 = ?

Now based on the law of conservation of mass:

Mass of B2S3 + mass C = mass of B + mass of CS2

Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5

Mass of B2S3 = 15.7 * 10^4 g

4 0

2 years ago

In an experiment, 0.42 mol of co and 0.42 mol of h2 were placed in a 1.00-l reaction vessel. at equilibrium, there were 0.29 mol

Likurg_2 [28]

To determine the Keq, we need the chemical reaction in the system. In this case it would be:

CO + 2H2 = CH3OH

The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:

CO H2 CH3OH
I .42 .42 0
C -0.13 -2(0.13) 0.13
-----------------------------------------------
E = .29 0.16 0.13

Therefore,

Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51

4 0

2 years ago