Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:
The mixture would contain
if 
undergoes no hydrolysis; the solution is of volume 
after the mixing. The two species would thus be of concentration 
and 
, respectively.
Construct a RICE table for the hydrolysis of under a basic aqueous environment (with a negligible hydronium concentration.)
The question supplied the <em>acid</em> dissociation constant 
for acetic acid 
; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant 
for its conjugate base, . The following relationship relates the two quantities:
... where the water self-ionization constant 
under standard conditions. Thus 
. By the definition of :
![[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BHAc%7D%20%28aq%29%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5D%20%2F%20%5B%5Ctext%7BAc%7D%5E%7B-%7D%20%28aq%29%20%5D%20%3D%20K_b%20%3D%20%2010%5E%7B-pK_%7Bb%7D%7D%20)
![[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%200.30%20%2Bx%20%5Capprox%200.30%20%5C%3B%20%5Ctext%7BM%7D%20)
![pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%20pK_%7Bw%7D%20-%20pOH%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%7B0.30%7D%20%3D%2013.5%20)
The answer to the question is 300meters.
This is called
Nuclear fusion
Example is two hydrogen atoms fuse to form a helium atom
Where some of the atoms of hydrogen turn in to energy
So the answer is A
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A
of
and
of
