Answer:
4,320 minutes
Explanation: There are 24 hours in a day and 60 minutes make up an hour, so just do the math there and it gives you your answer!
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.
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The ions of Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
The periodic table is a systematic arrangement of elements in order of their atomic numbers into a set of 8 columns each called groups and a set of 7 rows each called a period.
Elements are arranged in different groups according to the number of Valence electrons they have.
- For instance, elements in the group I of the periodic table are highly electropositive and as such are highly reactive.
The same is evident in group 7 elements are highly electronegative and have high electron affinity and as such are unstable and reactive.
- However, Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
Consequently, the <em>Noble gases ion</em> has a stable Valence electron configuration.
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D. With the same number of protons and different number of neutrons.
From the calculation, the pH of the solution after dilution is 3.
<h3>What is the pH?</h3>
The pH is the hydrogen ion concentration of the solution. Now we know that;
C1 = 0.010 m
V1 = 10.0 ml
V2 = 10.0 ml + 100.0 ml = 110 ml
C2 = ?
C1V1 = C2V2
C2 = C1V1 /V2
C2 = 0.010 m * 10.0 ml / 110 ml
C2 = 0.00091 M
pH = -log[0.00091 M]
pH = 3
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