Question

PLEASE HELP ME CHEMISTRY NERDS

1. A 0.4688 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.8165 g.

What is the mass percentage of bromine in the original compound?

2. A student determines the chromium(II) content of a solution by first precipitating it as chromium(II) hydroxide, and then decomposing the hydroxide to chromium(II) oxide by heating.

How many grams of chromium(II) oxide should the student obtain if his solution contains 32.0 mL of 0.503 M chromium(II) nitrate?

3. The iodide ion concentration in a solution may be determined by the precipitation of silver iodide.
Ag+(aq) + I-(aq) ----> AgI(s)
A student finds that 24.79 mL of 0.7220 M silver nitrate is needed to precipitate all of the iodide ion in a 25.00-mL sample of an unknown.

What is the molarity of the iodide ion in the student's unknown?

Answer 1

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Step-by-step explanation:

.

Ag+(aq) + I-(aq) ----> AgI(s)

A student finds that 24.79 mL of 0.7220 M silver nitrate is needed to precipitate all of the iodide ion in a 25.00-mL sample of an unknown.

What is the molarity of the iodide ion in the student's unknown?.

Ag+(aq) + I-(aq) ----> AgI(s)

A student finds that 24.79 mL of 0.7220 M silver nitrate is needed to precipitate all of the iodide ion in a 25.00-mL sample of an unknown.

What is the molarity of the iodide ion in the student's unknown?.

Ag+(aq) + I-(aq) ----> AgI(s)

A student finds that 24.79 mL of 0.7220 M silver nitrate is needed to precipitate all of the iodide ion in a 25.00-mL sample of an unknown.

What is the molarity of the iodide ion in the student's unknown?.

Ag+(aq) + I-(aq) ----> AgI(s)

A student finds that 24.79 mL of 0.7220 M silver nitrate is needed to precipitate all of the iodide ion in a 25.00-mL sample of an unknown.

What is the molarity of the iodide ion in the student's unknown?