Hydrochloric Acid (HCl) is a strong acid when it is present in the concentrated form but when it is dissolved in water (
) the atoms of this compound dissociates into its respective ions as shown below:
When we add HCl to any complex in its concentrated form the complex does not react at all but when its diluted to 6M and is kept for many hours, the complex reacts slowly. For eg:
![[Co(NH_3)_6]^3^+(aq.) + HCl(aq.) \rightarrow [Co(NH_3)_5Cl]^2^+(aq.) + NH_4^+ (aq.)](https://tex.z-dn.net/?f=%5BCo%28NH_3%29_6%5D%5E3%5E%2B%28aq.%29%20%2B%20HCl%28aq.%29%20%5Crightarrow%20%20%5BCo%28NH_3%29_5Cl%5D%5E2%5E%2B%28aq.%29%20%2B%20NH_4%5E%2B%20%28aq.%29)
As seen from the above reaction it can be seen the positive charge on the complex is reduced by 1 unit because one 
ion gets attached to the centre metal atom, therefore we can conclude that the charge on complex gets reduced by 1 unit when HCl reacts with the complex.
Answer: a) if a proton transforms to a neutron, a positron is produced
B) when an neutron transforms into a proton, an electron is produced
Explanation:
The both are nuclear decay processes which produce a neutrino and tremendous energy. The conversion of protons to neutrons is an energetically difficult process. However, the conversion of neutrons to electrons is commonly called beta decay in nuclear physics.
Heat the sulfuric acid solution and adding copper carbonate in it.
<h3>Improvement require in this experiment</h3>
The method could be improved by heating sulfuric acid solution and add copper carbonate into the solution of sulfuric acid. Add the copper carbonate until it is present in excess amount. After that filter the extra amount of copper carbonate so in this way the blue copper sulfate crystals are produced.
Learn more about copper here: brainly.com/question/3157958
B 0.146m
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Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:
Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)
From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,
Let's solve this for x. Multiply both sides by 0.12
taking square root to both sides:
Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
