Three. They belong to the bromide ion.
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Ka = [H+].[Br-] / [HBr] However the value of this expression is very high, because HBr is a STRONG acid, meaning that much more than 99.9% of the HBr molecules in water are protolized (ionized), making [H+] and [Br-] equal to the original (added) HBr amount, and the [HBr]-value nearly zero.
There is no conjugate) base coupled to bromide, Br-, because this Br- can NOT donate (by protolysing) a proton (H+) in water.However Br- itself is the very, very weakest base of the (very, very) strong conjugate acid HBr.This is the only possible conjugate acid/base pair:HBr/Br-
In the reaction, HBr donates a proton (H+) to H2O, making HBr the acid and H2O the base. The resulting products are Br- (conjugate base of HBr) and H3O+ (conjugate acid of H2O).
There is one hydrogen atom and one bromine atom in one molecule of HBr.
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Two lone pair on the central selenium and three lone pairs on each chlorine. So total of eight lone pairs.
Ka = [H+].[Br-] / [HBr] However the value of this expression is very high, because HBr is a STRONG acid, meaning that much more than 99.9% of the HBr molecules in water are protolized (ionized), making [H+] and [Br-] equal to the original (added) HBr amount, and the [HBr]-value nearly zero.
you have 5 binded elektron pairs, and one free pair
There is no conjugate) base coupled to bromide, Br-, because this Br- can NOT donate (by protolysing) a proton (H+) in water.However Br- itself is the very, very weakest base of the (very, very) strong conjugate acid HBr.This is the only possible conjugate acid/base pair:HBr/Br-
H-Br
In the reaction, HBr donates a proton (H+) to H2O, making HBr the acid and H2O the base. The resulting products are Br- (conjugate base of HBr) and H3O+ (conjugate acid of H2O).
There is one hydrogen atom and one bromine atom in one molecule of HBr.
Polar!
No, HBr is not an element. It is a compound made up of the elements hydrogen (H) and bromine (Br).
The equation for the reaction between hydrobromic acid (HBr) and water (H2O) can be represented as: HBr + H2O → H3O+ + Br-. This reaction involves the transfer of a proton from HBr to water, resulting in the formation of hydronium ion (H3O+) and bromide ion (Br-).
HCl (aq) + H2O (L) ---------> H3O+ (aq) + Cl- (aq)