Based on <span> Henderson - Hasselbach equation, we have
pH = pka + log </span>
Given: pH = 5.28 and [Acid] = 0.05 m,
∴ pKa = 5.28 - log (0.05) = 6.58
∴ Ka = 2.624 X
Thus, ionization constant of boric acid is 2.624 X
Answer is: A) The shape.
Name of this organic chemical compound is cyclohexane.
Cyclohexane is cycloalkane (the monocyclic saturated hydrocarbons).
Cyclohexane has single covalent bond and carbons have sp3 hybridization as other alkanes (for example propane).
Propane is alkane, organic compound. Carbons in propane have sp3 hybridization (carbon’s 2s and three 2p orbitals combine into four identical sp3 orbitals). Orbitals in sp3 hybridization have a tetrahedral arrangement and form single (sigma) bonds.
The First Two Are Correct Since They Give Off Light
Answer:
The names of the electron shell were given by a spectroscopist named Charles G Barkla. He named the innermost shell has k shell because he noticed that the X-rays emitted two types energies. These energies were named as type A that is higher energy X-ray and type B that is lower energy X-ray. Later these energiers were named with different alphabets. He noticed that K type X-rays emitted the highest energy. Therefore, he named the innermost shell as the K shell.
Explanation:
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Answer:The structures of 3-brominated products are available in attachment. Kindly find in attachment.
Explanation:
1-ethyl-4-methylbenzene undergoes a radical substitution reaction when it is treated with NBS(N-bromosuccinimide).
There are 3 products which are produced in the reaction.
There are 2 positions available where bromination can occur one at 1 position and other at 4 position.
There is a ethyl group present at 1-position and a methyl group is present at 4-position.
At the 1-position where ethyl group is present 1-phenylethyl radical is generated on irradiation with UV-light. Now since this 1-phenylethyl radical is planar and a chiral centre as all groups attached to the carbon center is different hence two products of bromination can occur from this position. One from below the plane and other from above the plane. These 2 products can form with equal probability.
In one product the Bromine radical can combine with 1-phenylethyl radical form above the plane and in other product bromine radical can combine with 1-phenyl radical form below the plane.
Hence 1-phenylethyl radical would lead to products which would be 2 stereiosomers and would be known as enantiomers (mirror images).
Radical generated at 4 position that is at methyl position would be a benzyl radical and this would also be planar but since it is not a chiral center hence both the sides would be equivalent so only one product would be generated.
Kindly find in attachment for the structures of the products and reactants.