A charming GT professor finds a bottle of mineral water. The label is torn off except for the analysis below in ppm: Na+ 5.50 K+ 3.90 Ca2+ 29.40 Mg2+ 5.10 SO4-- 31.40 Cl- 2.84 HCO3-91.00 pH 6.76 Use the following molecular weights Ca2+ 40.1 g/mole, Na+ 23 g/mole, Mg2+ 24.3 g/mole, K+ 39.1 g/mole, SO42-96.1 g/mole, Cl- 35.5 g/mole, HCO3- 61.0 g/mole. a) compute the charge balance b) calculate the ionic strength c) calculate the activity coefficients for H+, Ca2+, HCO3- and CO32-

A charming GT professor finds a bottle of mineral water. The label is torn off except for the analysis below in ppm: Na+ 5.50 K+ 3.90 Ca2+ 29.40 Mg2+ 5.10 SO4-- 31.40 Cl- 2.84 HCO3-91.00 pH 6.76 Use the following molecular weights Ca2+ 40.1 g/mole, Na+ 23 g/mole, Mg2+ 24.3 g/mole, K+ 39.1 g/mole, SO42-96.1 g/mole, Cl- 35.5 g/mole, HCO3- 61.0 g/mole. a) compute the charge balance b) calculate the ionic strength c) calculate the activity coefficients for H+, Ca2+, HCO3- and CO32-

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter3: Chemical Reactions

Section: Chapter Questions

Problem 87QRT: You need 300. mL of 0.500-M K2Cr2O7. Which method is best to prepare this solution? Explain your...

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Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A 0.378-m solution is prepared by dissolving 38.4 g sodium diatrizoate (NaDTZ) in.l.60 102 mL water at 3 1.2C (the density of water at 31.2C is 0.995 g/cm3). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at 31.2C is 34.1 torr?
You need 300. mL of 0.500-M K2Cr2O7. Which method is best to prepare this solution? Explain your choice. (a) Dilute 250. mL of 0.600-M K2Cr2O7 to 300. mL. (b) Add 50.0 mL water to 250. mL of 0.250-M K2Cr2O7. (c) Dilute 125 mL of 1.00-M K2Cr2O7 to 300. mL. (d) Add 30.0 mL of 1.50-M K2Cr2O7 to 270. mL of water.
how do you calculate per serving Zn value for kidney beans compare it with the reported value fine powder beans: 3.3011 g nutrtional value per 100g (3.5 oz): zinc 10%. 1.07 mg y=0.0037x+0.2727 0=0.0037x+0.2727 x=-73.702 = 73.702 Diluted of unknown solution: 73.702 ppb x 400 (dilution factor) = 29,480 ppb 29,480 ug/L x 0.010 = 2948 ug = 2.94 mg in ca 4 g beans
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample.
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample. The stated should be 104.5, how??
C. Electrolytes in Body Fluids 1. Type of IV Solution 5% Isolute P (mEa/L) Plasnalyte (mEg/L) Lactated Ringer's (mEg/L) Nal+ = 130 Klt = 4 Ca2+ = 3 Nal+ = 25 K+ = 20 Nal+ = 140 Kl+ = 5 2. Cations %3D %3D Mg+ = 3 Mg2+ = 3 CI- = 22 HPO- =3 Acl- = 23 CI- = 98 Glu'-= 23 Acl- = 27 3. Anions CI'- = 109 %3D HPO = 28 %3D %3D 4. Total Charge of Cations (+) 137 137 5. Total Charge of Anions (-) 6. Sum of the Charges Copyright 2014 Pearson Education, Inc.
IZVW @ ol:09 zain IQ li. How many millilitres of concentrated sulphuric acid, 99.0%, density 1.831 g/cm3, are required to prepare 0.6 liter of a 0.2 M solution? Note ( M.wt of sulphuric acid = 98.1 g/mol) * %3D 5.7 ml 18.4 ml 6.5 ml 17.5 ml prepare a standard solution of a potassium Chloride (KCI), 0.3 N in 500 ml of water. (atomic weight : K=19 , Cl=35 ) ? * 1.8 gm 8.1 gm 81 gm 0.81 gm docs.google.com a
Calculate the concentration of the solution: 2 g C12H22O11 / 100 g H2O using % by Mass, % by Volume, and Parts Per Million (ppm). Hint: mass 1 mL H2O = 1 gram1 L solution = 1000 mL
2. How many mL of a 0.250 M KCI solution must be diluted to 1.000 L so that the diluted solution (density = 1.00 g/mL ) is 400 ppm K* by weight ? (MW KCI = 74.55; AW K = 39.10)
Given the following: Solutions Flask 1 Flask 2 Flask 3 0.04 M KI 25.0 mL 25.0 mL 12.5 mL 1% starch 5.00mL 5.00mL 5.00mL 0.03 M Na2S2O3 5.50 mL 5.50 mL 5.50 mL 0.05 M H2SO4 5.00 mL 5.00mL 5.00mL 0.08 M NaCl 0 0 10.0 mL and Solutions Beaker 1 Beaker 2 Beaker 3 0.04 M H2O2 25.0 mL 12.5 mL 25.0 mL deioinized water 0 12.5 mL 0 Following the procedure Run 1 is when Beaker 1 is added to Flask 1 and the timer is started giving the time recorded for the run in the table below. This is repeated with Beaker 2 and Flask 2 for Run 2 and finally Beaker 3 and Flask 3 for Run 3: The results for these runs are given below: Run Time (seconds) 1 35 2 66 3 72 Calculate the following for each run: the initial molarities of I- and H2O2 (from the solutions tables) the rates of reaction [I2]/time (calculate the [I2] from [S2O32-] and the reaction: I2(aq) + 2 S2O32-(aq) → 2I-(aq) + S4O62-(aq) Choose the correct rate law for this reaction from your calculations.…
Give only typing answer with explanation and conclusion
How many grams of Ba(103)2 (487 g/mol) can be dissolved in 400 mL of water at 25°C. (Ksp of Ba(1O3)2 = 1.57 x 10-9) Ba(103)2 Ba2+ + 2103 O 0.298 g 0.122 g 0.221 g O 0.926 g O 0.143 g