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:
There is one single covalent bond between two carbon atoms.
Explanation:
We know that sharing of electrons form covalent bonds.
If we look upon K,L,M ,N shells of the carbon and hydrogen atoms.
We found that Hydrogen is having only 
electron in K shell.
And Carbon on the other hand is having 
electrons in K shell and 
electrons in L shell.
So carbon have valence electrons,and it can share bonds with any relevant atom to complete its octet.
And Hydrogen requires electron to complete its doublet.
Alkane general formula 
For ethane 
Carbon atom is shared by 
Hydrogen.
The remaining one electron 
of carbon will be shared with another carbon atom.
An image of the sharing of electrons attached below,
Hence we have only covalent bond between the two.
Answer:
Explanation: I think its 4.91 x 10^25. Im not very sure, i just multipled 1.15 mol by the molar mass of Cl 2, which was 70.9 g. Then I multiplied that by avogadro's number. sorry if im wrong
Answer:
The molarity of the solution: 0,27M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl
A solution molar--> moles of solute in 1 L of solution:
4 L-----1,09 mol NaCl
1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M
