2. Aqueous lead (II) nitrate reacts with aqueous potassium chloride and produces
lead (II) chloride and potassium nitrate. The reaction begins with 250.0 mL of
5.5 M solution of lead (II) nitrate and excess potassium chloride, how many grams of the
precipitate (solid) will formę

there all right good

explanation:

you need to have a picture of the atom that way we can answer it

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol