Write a balanced molecular equation for the reaction between solid aluminum hydroxide and a solution of hydrobromic acid. Include the state of each species.

Al(OH)₃(aq) + 3 HBr(aq) ⇒ AlBr₃(aq) + 3 H₂O(l)

Explanation:

Let's consider the unbalanced equation for the reaction between solid aluminum hydroxide and a solution of hydrobromic acid. This is a neutralization reaction so it forms a salt, aluminum bromide, and water.

Al(OH)₃(aq) + HBr(aq) ⇒ AlBr₃(aq) + H₂O(l)

We start balancing Br atoms by multiplying HBr by 3.

Al(OH)₃(aq) + 3 HBr(aq) ⇒ AlBr₃(aq) + H₂O(l)

To get the balanced equation, we must multiply H₂O by 3 as well.

Al(OH)₃(aq) + 3 HBr(aq) ⇒ AlBr₃(aq) + 3 H₂O(l)

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally