Increase the frequency to a higher frequency
Answer:
Growth of weeds and grasses
Explanation:
The part of the garden, probably left to rest by the Gardner, would start experiencing a legion of invasive plants. Giant weeds and grasses could be seen, and brambles and tree seedlings could also sneak in.
Some Gardeners would recommend the use of herbicides to restore order to an abandoned garden even though certain scientists are of the opinion that chemicals may not offer a sound solution to the problem.
CH3CH2OH + 3O2 = 2CO2 + 3H2O
Basically you do trial and error on both sides so they can be equal
Answer:
8.73
Explanation:
The concentration of acetic acid can be determined as follows:






Moles of
= 
=0.0090 moles
Moles of 
= 0.0090 moles
The equation for the reaction can be expressed as :
----->

Concentration of
ion = 
= 
= 0.052 M
Hydrolysis of
ion:
----->


⇒ 
= 
As K is so less, then x appears to be a very infinitesimal small number
0.052-x ≅ x





![[OH] = x =0.535*10^{-5}](https://tex.z-dn.net/?f=%5BOH%5D%20%3D%20x%20%3D0.535%2A10%5E%7B-5%7D)
![pOH = -log[OH^-]](https://tex.z-dn.net/?f=pOH%20%3D%20-log%5BOH%5E-%5D)
![pOH = -log[0.535*10^{-5}]](https://tex.z-dn.net/?f=pOH%20%3D%20-log%5B0.535%2A10%5E%7B-5%7D%5D)

pH = 14 - pOH
pH = 14 - 5.27
pH = 8.73
Hence, the pH of the titration mixture = 8.73