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:
The transition from lower energy level to higher energy level require a gain of energy.
Explanation:
When transition occur from lower energy level to higher energy level require a gain of energy. Electron could not jump unto higher energy level without gaining thew energy.
When electron jump into lower energy level from high energy level it loses the energy.
For example electron when jumped from 2nd to 3rd shell it gain energy and when in return back to 2nd shell from 3rd shell it loses energy.
The process is called excitation and de-excitation.
Excitation:
When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.
De-excitation:
When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.
Answer: A wave
Explanation:
Because it’s the one that’s cause the new medium to go between the two media.
Ca2+ would bond to any element in a 1 to 1 ratio that had an equal and opposite charge.
Neon is a noble gas, and doesn’t form bonds m
Carbon isn’t typically found in ion state, but if it did, it would likely by C4+
Flouring in ionic state is F1-, so you would need 2 flourines to cancel the 2+ charge of Calcium
Then the only option left would be Oxygen which, when in ion form is found be 2-
