A student has a mixture of salt (NaCl) and sugar (C12,H22O11). To determine the percent composition, the student measures out 5.84 g of the mixture and dissolves the sample in 100.0 mL of water. The student knows aqueous NaCl will react with aqueous silver nitrate (AgNO3) to form solid silver chloride (AgCl). The student determines the sugar will not react with silver nitrate. How many mL of 1.0 M AgNO3 will be required to precipitate 5.84 g of AgCl?

Explanation:

Every atom of Ag in AgNO₃ will be present in the AgCl precipitate. Thus, you can determine the amount of Ag in 5.84g of AgCl, and then the volume of AgNO₃ that contains that amount.

1. Ag in 5.84g of AgCl

You can set a proportion using the mass of Ag in one mole of AgCl:

2. Number of moles of AgNO₃ that contain 4.395 g of Ag

Set a proportion, too:

3. Volume of AgNO₃

Use the molarity of the AgNO₃ solution:

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

the answer is c.the valence of the atoms combine .