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
Sodium phosphate = Na₃PO₄
phosphorus triiodide = PI3
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One of the products of photosynthesis is carbon dioxide
The balanced chemical equation is written as:
<span>CsF(s) + XeF6(s) ------> CsXeF7(s)
We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:
11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
Answer:
The correct answer is - 800.
Explanation:
Given:
Total amount = ? or assume x
spend in buying birthday item = 3/4 of x
given to sister = 1/5 of x
remaining to mother = 40
solution:
the remaning amount = x- (3x/4+x/5) = 4=
=> x- 19x/20 = 40
=> x = 20*40
=> x = 800
thus, the correct answer is = 800
