Answer:
pKb = 10.96
Explanation:
Tartaric acid is a dyprotic acid. It reacts to water like this:
H₂Tart + H₂O ⇄ H₃O⁺ + HTart⁻ Ka1
HTart⁻ + H₂O ⇄ H₃O⁺ + Tart⁻² Ka2
When we anaylse the base, we have
Tart⁻² + H₂O ⇄ OH⁻ + HTart⁻ Kb1
HTart⁻ + H₂O ⇄ OH⁻ + H₂Tart Kb2
Remember that Ka1 . Kb2 = Kw, plus pKa1 + pKb2 = 14
Kb2 = Kw / Ka1 → 1×10⁻¹⁴ / 9.20×10⁻⁴ = 1.08×10⁻¹¹
so pKb = - log Kb2 → - log 1.08×10⁻¹¹ = 10.96
Answer:
The concentration of cyclopropane after 22.0 hour is 0.0457 M.
Explanation:
Conversion of cyclopropane into propene follows first order kinetics.
The integrated rate of first order kinetic is given by :
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= Initial concentration of reactant
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= final concentration of reactant after time t
k = rate constant of the reaction
We have :
Rate constant of the reaction = k = 
![[A_o]=0.150 M](https://tex.z-dn.net/?f=%5BA_o%5D%3D0.150%20M)
t = 22.0 hour
[A] =?
![[A]=0.150 M\times e^{-5.4\times 10^{-2} hour^{-1}\times 22.hour}](https://tex.z-dn.net/?f=%5BA%5D%3D0.150%20M%5Ctimes%20e%5E%7B-5.4%5Ctimes%2010%5E%7B-2%7D%20hour%5E%7B-1%7D%5Ctimes%2022.hour%7D)
![[A]=0.0457 M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0457%20M)
The concentration of cyclopropane after 22.0 hour is 0.0457 M.
Answer: This contains magnesium, Mg2+, and hydroxide, OH–
, ions. Each magnesium ion is +2 and
each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and the formula for magnesium
hydroxide is Mg(OH)2. The (OH)2 indicates there are two OH–
ions. In a formula unit of
Mg(OH)2, there are one magnesium ion and two hydroxide ions; or one magnesium, two
oxygen, and two hydrogen atoms. The subscript multiplies everything in ( )
hope that helped!!
The answer is more protons than electrons.
Answer:
The materials are opaque or crystalline from a client to the orientation and type of union between their atoms, forming two types of structures.
These two structures can be crystalline or amorphous.
In the case of being crystalline, these unions do not allow light to pass through the medium of the object or body of said compound, making it totally refract and giving the appearance of OPAQUE.
On the other hand, in those compounds that we call amorphous, the atoms are located in a different way that makes light pass through them, without absorbing or identifying any light beam, so they look transparent.
Explanation:
Example: A glass cup has an amorphous structure, while a porcelain or porcelain plate has a crystalline structure.
