Answer:
Explanation:
In the following reaction we have shown an example of aromatic substitution reaction .
C₆H₆ + RCl = C₆H₅R + HCl
This reaction takes place in the presence of catalyst like AlCl₃ which is a lewis acid .
First of all formation of carbocation is made as follows .
RCl + AlCl₃ = R⁺ + AlCl₄⁻
This R⁺ is carbocation which is also called electrophile . It attacks the ring to get attached with it .
C₆H₆ + R⁺ = C₆H₅R⁺H.
The complex formed is unstable , though it is stabilized by resonance effect . In the last step H⁺ is kicked out of the ring . The driving force that does it is the steric hindrance due to presence of two adjacent group of H and R⁺ at the same place . Second driving force is attack by the base AlCl₄⁻ that had been formed earlier . It acts as base and it extracts proton ( H⁺ ) from the ring .
C₆H₅R⁺H + AlCl₄⁻ = C₆H₆ + AlCl₃ + HCl .
The formation of a stable product C₆H₆ also drives the reaction to form this product .
Chemical or molecular formulas? Symbols from the periodic table are used to show which atoms are present in a molecule and subscripts show the number of atoms and of each type in the molecule.
Answer:
[H₃O⁺] = 1.4 × 10⁻⁹ M.
Explanation:
NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.
NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?
Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.
Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:
.
The 
value for ammonia is small. The value of x will be so small that at equilibrium, 
and 
.
![\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BK%7D_b%20%3D%20%5Cfrac%7B%5B%7B%5Ctext%7BNH%7D_4%7D%5E%7B%2B%7D%5D%5Ccdot%20%5B%7B%5Ctext%7BOH%7D%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BNH%7D_3%5D%7D%20%5Capprox%20%5Cfrac%7B2.5%5C%3B%28x%20%2B%201.0%5Ctimes%2010%5E%7B-7%7D%29%7D%7B1.0%7D)
.
.
![\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20x%2B1.0%5Ctimes%2010%5E%7B-7%7D%20%3D%201.8%5Ctimes%2010%5E%7B-5%7D%20%2F%5Cleft%28%5Cfrac%7B2.5%7D%7B1.0%7D%5Cright%29%20%3D%207.2%5Ctimes%2010%5E%7B-6%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7BL%7D%5E%7B-%7D)
.
Again, 
at room temperature.
![\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BH%7D_3%5Ctext%7BO%7D%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B%5Ctext%7BK%7D_w%7D%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%7D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-14%7D%7D%7B7.2%5Ctimes%2010%5E%7B-6%7D%7D%20%3D%201.4%5Ctimes%2010%5E%7B-9%7D%20%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7BL%7D%5E%7B-1%7D)
Answer:
The statement correctly predicting and explaining the chemical reactivity of two metals is given below -Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons
Explanation:
Answer:
C1V1=C2V2
C1 is 2.0mol/l
V1=?
C2=.4mol/L
V2=100ml or for this 0.1L
V1 is 20ml
Best way to prepare this is to measure out 20ml of the 2 molar solution and add 80mL to it to get to 100mL
Explanation:
