The FD&C blue (Molar Mass 792.86 g/mole) is a very intensely colored blue dye and only dilute solutions are needed. The stock solution provided was made by dissolving 0.14 g in water and making the solution to a total volume of 100 mL 1. What is the concentration (M) of the stock solution? In this experiment you need to take 1.00 mL of this stock solution and dilute it to 100 mL in a volumetric flask. What will be the concentration of diluted dye solution? 2. The other reagent used in this experiment is bleach. Bleach is simply a 6.15% solution of sodium hypochlorite, NaCIO, dissolved in water (6.15 g of NaCIO in 100 g solution). Determine the molarity of the bleach solution. The density of bleach is 1.10 g/mL. 3. For the first experiment, you will be mixing 1.00 mL of the dye solution, 1.50 mL of water, and 0.50 mL of bleach in a cuvet. Using the concentrations you calculated in questions 2 & 3, determine the concentrations of the FD&C blue and CIO that are initially present in the cuvet when they have all been mixed together. Note that the final total volume is 3.00 mL. 4. 5. Describe how Beer's Law can be used to determine the concentration of a solution

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

so she was walking and she stepped on warm jeffy , uh uh uhh dadye i just pooped my pants

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