A student with an unknown code of Q3A conducts a qualitative analysis of her unknown. Her results indicated the formation of a yellow precipitate and a grey precipitate. From her results, what ions can you confirm were not in her unknown sample?

Grey precipitate implies the presence of silver ions

Yellow precipitate implies the presence of lead II ions

Explanation:

Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.

In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

bcde is the incorrect answer it is a