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:
964ug
Explanation:
The problem here involves converting from one unit to another.
We are to convert from ounces to micrograms.
1ug = 1 x 10⁻⁶g
1oz = 28.35g
So we first convert to grams from oz then take to ug:
Solving:
1oz = 28.35g
3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g
So;
1 x 10⁻⁶g = 1ug
9.64 x 10⁻⁴g will give 
= 9.64 x 10²ug or 964ug
Answer:
Zr (Zirconium)
Explanation:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d2
Answer:
Explanation:
It can be determined by measuring the Ph. D is incorrect.
C: is wrong because if you are making something acidic, you are increasing the H+
B: is the correct answer.
A: pH decreases. H+ increases which makes the Ph decrease. It is an oddity of the formula that makes this happen.
