One difficulty encountered in precipitation titration is that it is hard to determine the exact end point of its reaction.
Precipitation titration is a titration in which a reaction occurs from the analyte and titrant to form an insoluble precipitate.
With the use of silver for the titrations, (argentometric) we are able to develop many precipitation reactions.
The precipitation titrimetry methods with the use of argentometry includes
• Mohr’s Method
• Fajan’s Method
• Volhard’s Method
Difficulties encountered in precipitation titration includes
- Getting the exact end point is hard.
- it is a very slow titration method.
- it includes periods of filtration and cooling thereby reducing the reactions available for this type of titration.
See more on Precipitation: brainly.com/question/20628792
Answer: A yellowy-green dense gas with a choking smell. Chlorine kills bacteria – it is a disinfectant. It is used to treat drinking water and swimming pool water. It is also used to make hundreds of consumer products from paper to paints, and from textiles to insecticides.
Explanation: hope this helps u! HAppy Holidays!
Answer:
Your answer is correct based on what I remember from AP Chemistry
Explanation:
Answer:
19.264×
atoms are present in 3.2 moles of carbon.
Explanation:
It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.
So, as 1 mole of any element = Avagadro's number of atoms = 6.02× atoms
It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.
As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.
1 mole of carbon = 6.02 × atoms
Then, 3.20 moles of carbon = 3.20 × 6.02 × atoms
Thus, 19.264× atoms are present in 3.2 moles of carbon.
Answer:
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v
Explanation:
