Answer:
- <em>The pH of the solution is </em><u><em>7</em></u>
Explanation:
<em>The pH</em> is a measure of the acidity of the solutions. It is defined as the negative logarithm of the molar concentration of hydrogen ions (H⁺).
<em>The hydrogen ion concentration of this solution is 1 × 10⁻⁷ M.</em>
Hence:
- pH = - log (1 × 10⁻⁷) = - (-7) = 7
This pH corresponds to a neutral solution (neither acid nor alkaline).
You should remember this relation bwtween pH and acidity/alkaliinity:
- Low pH (0.0 or close) corresponds to strong acids
- HIgh pH (14.0 or close) corresponds to strong bases
- Acids have pH between 0.0 and 7.0
- Bases have pH between 7.0 and 14.0
Answer:
sorry
Explanation:
i'm not a chemistry student
but let me try
it is because the neutron of an atom is -
why the proton and electron is +and- (positive and negative)
so + and - = 0
Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
That would be evaporation.
Hope this helped!! xx