Answer:
pOH = 8.19
Acidic solution (pH < 7.0)
Explanation:
The pOH of the solution is defined as:
pOH = - log [OH⁻]
Since we have [OH⁻] = 6.5 x 10⁻⁹ M, we calculate the pOH as follows:
pOH = - log (6.5 x 10⁻⁹) = 8.19
To know if the solution is acidic, neutral or basic, we have to calculate the pH from the value of pOH:
pH + pOH = 14
⇒ pH = 14 - pOH = 14 - 8.19 = 5.81
The solution is acidic because pH < 7.0.
The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell.
The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.
Answer:
It's a rapid motion and collisions of molecules with the walls of the container that causes pressure.
Answer:
(3S)-2-chloro-2,3-dimethylpentane is produced exclusively.
Explanation:
Electrophilic addition to (3S)-2,3-dimethylpent-1-ene proceeds through a carbocationic intermediate.
In the first step, 
adds onto double bond to produce more stable tertiary carbocation. (protonation)
In the second step, 
adds onto carbocation to produce (3S)-2-chloro-2,3-dimethylpentane exclusively.(nucleophilic addition)
So, option (d) is correct.
Answer:
See explanation and image attached for details
Explanation:
The reaction involves the heterolytic fission of the Br-Br bond in the bromine molecule to yield a bromine cation which attacks the but-1-ene to form a cyclic intermediate called the brominium ion. The bromine anion must now attack from the opposite face of the brominium ion due to steric clashes to form a product of a 1,2-dibromoalkane having the anti- stereochemistry.
