A chemist prepares a solution of aluminum sulfite by measuring out 109 umol of aluminum sulfite into a 350 mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mmol/L of the chemist's aluminum sulfite solution. Be sure your answer has the correct number of significant digits.

0.311 mmol/L

Explanation:

109  μmol = 109*10^(-6) mol

109*10^(-6) mol = 109*10^(-6) mol*(10^3 mmol/1mol) = 109*10^(-3) mmol =

=0.109 mmol

350 mL = 0.350 L

0.109 mmol/0.350 L = 0.311 mmol/L

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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