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Chapter 27: Q3P (page 767)

Why is high relative supersaturation undesirable in a gravimetric precipitation?

Short Answer

Expert verified

The high relative supersaturation undesirable in a gravimetric precipitation due to the fast nucleation- suspension of colloids would be higher would isn’t desirable.

Step by step solution

01

Definition

It is an analytical technique that uses a precipitation reaction to separate ions from a solution. The chemical that is added to cause the precipitation is called the precipitant or precipitating agent

02

Formation of Aggregrates

During the process of nucleation in gravimetric precipitation process, molecules are closer to one another in a solution which leads to the formation of aggregrates.

03

Relative supersaturation is undesirable

This process appear to be faster in supersaturated solutions, and due to fast nucleation, suspension of colloids would be higher would isn’t desirable.

Therefore, in gravimetric precipitation high relative supersaturation is undesirable.

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Most popular questions from this chapter

An organic compound with a formula mass of 417g/molwas analyzed for ethoxyl $({CH}_{y} {CH}_{2} O -)$ groups by the reactions

${ROCH}_{2} {CH}_{3} + HI \to ROH + {CH}_{3} {CH}_{2} I {CH}_{3} {CH}_{2} I + {Ag}^{+} + {OH}^{-} \to AgI (s) + {CH}_{3} {CH}_{2} OH$

A 25.42-mg sample of compound produced How many ethoxyl groups are there in each molecule?

Combustion analysis of a compound known to contain justC,H,N, and Odemonstrated that it is 46.21wt%C,9.02wt% H, 13.74wt N and, by difference, 100 - 46.21 - 9.02 -13.74 = 31.03% O. This means that of unknown would contain46.21gof C,9.02g ofH, and so on. Find the atomic ratio C : H : N : Oand express it as the lowest reasonable integer ratio.

Why is sample dropped into the preheated furnace before the oxygen concentration reaches its peak in Figure 27-9?

Finely ground mineral (0.6324g)was dissolved in 25 mLof boiling 4M HCland diluted with $175 mL H_{2} O$ containing two drops of methyl red indicator. The solution was heated to $100^{o} C, and 50 mL$ of warm solution containing $2.0 g ({NH}_{4})_{2} C_{2} O_{4}$ were slowly added to precipitate ${CaC}_{2} O_{4} . Then 6 {MNH}_{3}$ was added until the indicator changed from red to yellow, showing that the liquid was neutral or slightly basic. After slow cooling for 1 h, the liquid was decanted and the solid transferred to a filter crucible and washed with cold $10.1 wt % ({NH}_{4})_{2} C_{2} O_{4}$ solution five times until no ${Cl}^{-}$ was detected in the filtrate upon addition of ${AgNO}_{3}$ solution. The crucible was dried at 1 h and then at $105 ° C$ in a furnace for 2 h.

${Ca}^{2 +} + C_{2} O_{4}^{2 -} \overset{105 ° C}{\to} {CaC}_{2} O_{4} + H_{2} O (s) \overset{500^{o} C}{\to} {CaCO}_{3} (s)$

FM 40.078 18.5467 g

The mass of the empty crucible was 18.2311 g and the mass of the crucible with ${CaCO}_{3}$ was 18.5467 g .

(a) Find the wt% Ca in the mineral.

(b) Why is the unknown solution heated to boiling and the precipitant solution, $({NH}_{4})_{2} C_{2} O_{4}$ also heated before slowly mixing the two?

(c) What is the purpose of washing the precipitate with0.1wt% $({NH}_{4})_{2} C_{2} O_{4}$ ?

(d) What is the purpose of testing the filtrate with ${AgNO}_{3}$ solution?

The thermogravimetric trace on the next page shows mass loss by $Y_{2} (OH)_{5} Cl \cdot {xH}_{2} O$ upon heating. In the first step, waters of hydration are lost to give ~8.1% mass loss. After a decomposition step,19.2% of the original mass is lost. Finally, the composition stabilizes at $Y_{2} O_{3}$ above $800^{9} C$ .

(a) Find x in the formula $Y_{2} (OH)_{5} Cl \cdot {xH}_{2} O .$ Because the 8.1% mass loss is not accurately defined in the experiment, use the 31.8% total mass loss for your calculation.

(b) Suggest a formula for the material remaining at the 19.2% plateau. Be sure that the charges of all ions in your formula sum to 0 . The cation is $Y^{3 +}$ .

See all solutions

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