you have CH3CH2CH2OH (propanol) + HBR = first you alcohol get protnated to good leaving group and to lower base which water . so you have CH3CH2CH2HO+ Br-. Then when water leaves you have final product of CH3CH2CH2Br + HOH.
1. CH3CH2CH3OH +HBR ( What you started with)
2. CH3CH2CH2HOH +Br- ( reaction after protnation of alcohol to good leaving group).
3. CH3CH2CH2BR + HOH ( final product primary haloalkane and water).
The balanced chemical reaction between methylamine (CH3NH2) and hydrobromic acid (HBr) is: CH3NH2 + HBr -> CH3NH3+ Br-
The neutralization reaction between HBr (hydrobromic acid) and Sr(OH)2 (strontium hydroxide) produces water (H2O) and strontium bromide (SrBr2) as products. The balanced chemical equation for this reaction is 2HBr + Sr(OH)2 → 2H2O + SrBr2.
CdBr2 (cadmium bromide) and H2(g) are produced in the reaction between solid Cd and aqueous HBr.
The products of the reaction between hydrogen bromide (HBr) and sodium hydroxide (NaOH) are sodium bromide (NaBr) and water (H2O). This is a neutralization reaction where the acid (HBr) reacts with the base (NaOH) to form a salt (NaBr) and water.
The reaction between HBr and KOH is a 1:1 ratio. This means that the moles of HBr present in the solution will be equal to the moles of KOH used in the neutralization reaction. Using this information and the volume and concentration of KOH used, you can calculate the concentration of the HBr solution.
The balanced chemical reaction between methylamine (CH3NH2) and hydrobromic acid (HBr) is: CH3NH2 + HBr -> CH3NH3+ Br-
This equation is:HBr + LiOH = LiBr + H2O
When bromine reacts with water, it forms hydrobromic acid (HBr) and hypobromous acid (HOBr). The overall reaction can be represented as: Br2 + H2O → HBr + HOBr. This reaction is reversible and depends on the pH and conditions of the solution.
The neutralization reaction between HBr (hydrobromic acid) and Sr(OH)2 (strontium hydroxide) produces water (H2O) and strontium bromide (SrBr2) as products. The balanced chemical equation for this reaction is 2HBr + Sr(OH)2 → 2H2O + SrBr2.
CH4 + Br2 --> CH3Br + HBr
CdBr2 (cadmium bromide) and H2(g) are produced in the reaction between solid Cd and aqueous HBr.
The balanced equation for the reaction between ammonia (NH3) and hydrobromic acid (HBr) is: NH3 + HBr -> NH4Br
The products of the reaction between hydrogen bromide (HBr) and sodium hydroxide (NaOH) are sodium bromide (NaBr) and water (H2O). This is a neutralization reaction where the acid (HBr) reacts with the base (NaOH) to form a salt (NaBr) and water.
The reaction between HBr and KOH is a 1:1 ratio. This means that the moles of HBr present in the solution will be equal to the moles of KOH used in the neutralization reaction. Using this information and the volume and concentration of KOH used, you can calculate the concentration of the HBr solution.
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).
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-).
The equation you mentioned is: KOH + HBr → KBr + H2O. This is a neutralization reaction between potassium hydroxide and hydrobromic acid to form potassium bromide and water.