The reaction of iron (III) metal with a solution of copper (II) sulfate releases iron ions into the solution through a single displacement reaction. 1. What mass of iron (in grams) is required to remove all the copper ions from a 150mL of 0.100 mol/L solution of copper (II) sulfate?
2. Determine the moles of iron ions produced in this reaction.
3. Name a soluble compound that could be added to precipitate all of the iron ions from the solution.
4. What mass of the soluble compound from part (c) is required to precipitate all of the iron ions you determined in part (b)? Give your final answer in grams.
(HINT: determine a new chemical reaction with your soluble compound Fe2(SO4)3. Then, use the moles of Fe2(SO4)3 calculated in part (b).)

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.

what is the question?

definition

                    entropy is a measure of the randomness or the extent of disorder of a chemical process.

                      enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

                        entropy is measured in joule per kelvin (jk⁻¹)   whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

                            entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

                              in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole