The following set of chloride analyses on separate aliquots of pooled serum sample were reported 107, 103, 114, 106 meq /L. One value appears suspect Determine if it can be ascribed to accidental error, at the 95% CL.
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- 1. What is the best answer to the following expression? A) B) C) D) (12.125 +0.530 +71.4) 84.055 84.06 84.1 84.0 84 E) What is the mass of a 55.5-mL sample of concentrated sulfuric acid (density 1.84 g/mL)? A) 30.2 g B) 40.2 g C) 82.0 g D) 92.0 g E) 102 g 3. The strongest intermolecular interactions between ethyl alcohol (CH CH OH) molecules arise from Dipole-dipole forces London dispersion forces A) C) D) E) B) Hydrogen bonding Covalent bonding Ion-dipole forces Which of the following is an example of a chemical change? water boiling ice melting 4. B) C) natural gas burning అంత - शभ D) alcohol evaporating E) iodine vaporizing ఆ A) 2.Aktiv Chemistr X b Success Confi x < + C app.101edu.co Introducing Ch x moles in Nal scx b Answered: Wh x b Answered: A 0 × b Answered: Hoxb Answered: As x Time's Up! What quantity in moles of Nal are there in 175.0 mL of 0.420 M Nal? 1 4 7 +/- 5 New Tab 2 3 8 0.420 mol 6 9 G 0 x + Submit X C x 100Calculate the molarity of caffeine in a 12 oz cola drink containing 31.1 mg of caffeine, C3H10N402. (1oz 29.6 mL) Submit Answer Tries 0/99 This discussion is closed. Send Feedback
- Practice Data For Lab Practical Data Set 1 • Stock KOH solution: 5.00M • Make 150.0 mL of 0.250M KOH fom the stock solution Show calculation Vl=250M(150.0mL.)/5 M → V1=7.50mL • Mass of pure H;C,0.*2H;0 standard that will use about 15.0mL of 0 250M KOH= %3D 05 126/2-63 → mass/63-25(15) → Mass 0.05g Standardization of KOH Mass of Oxalic Acid Dihydrate Trial 1 0.2360 Trial 2 0.2371 Trial 3 0.2364 (g) Initial Volume of 0.00 KOH (mL) Final Volume of KOH (mL) 14.89 29.92 14.89 29.92 44.95 Calculate the molarity of KOH for each trial, the average concentration of KOH and the precision of the three trials Trial 1 | I just need help with the 3rd question, calculating the molarity.Question 15 Waht is the composition of the reverse phase chromatography? O polar stationary phase - polar mobile phase O non- polar stationary phase - polar mobile phase O non-polar stationary phase - non polar mobile phase polar stationary phase - non polar mobile phaseThe following set of chloride on seperate all quots Of a pette pooled Serum obtaned from SMu chemistry lab here reporterd by (43) students : (03:106; (07 and 114 meqlL. analysis One value qppears suspect. Help these Students to determine if it can be ascribed to the acudental 95% Confidence level. error ot
- Analysis of a blood sample for inorganic phosphate gives (4.5, 4.20, 3.6, 4.17, 3.88) mg% phosphate. The value of Standard deviation = +- 0.573 a) True O b) False OQuantity Example Trial 1 Trial 2 Trial 3 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V; 12.90 mL 14.22 mL 13.71 mL 13.31 mL Vep = VNaoH added = V, - Vi 12.70 mL 13.02 mL 13.19 mL 13.16 mL Vep = VNaOH in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = M X VNAOH 'NaOH moles AA moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M 0.833 M of AA Average molarity for 3 trials => 0.833 M22:12 Question 23 of 30 Submit How many grams of a solution that is 5.2 % sucrose by mass are needed to obtain 18 g of sucrose X STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 18 100 g solution 3.5 6.022 x 1023 %m sucrose 0.052 5.2 g sucrose 94 350 0.0094 0.94 1 Tap here or pull up for additional resources
- What is the concentration in mg/m3 of a solvent detected at 0.15% in aatmospheric analysis of confined space whose molecular weight is 78.11 g/mol? Assume NTP. Please explain, correct answer and typed answer only, no copy paste ( no AI generated answer)Eagall 5olved its volume(4L)His focus() Nagl from sodiunchloride it was related by adding one liter of water, what molarisolutions output? Naz 23 cl=35-5Exercises S-A Detection limit In spectrophotometry, we measure the con- centration of analyte by its absorbance of light. A low-concentration sample was prepared, and nine replicate measurements gave ab- sorbances of 0.004 7, 0.005 4, 0.006 2, 0.006 0, 0.004 6, 0.005 6, 0.005 2, 0.004 4, and 0.005 8. Nine reagent blanks gave values of /0.000 6, 0.001 2, 0.002 2, 0.000 5, 0.001 6, 0.000 8, 0.001 7, 0.001 0, and 0.001 1. (a) Find the absorbance detection limit with Equation 5-3. (b) The calibration curve is a graph of absorbance versus concen- tration. Absorbance is a dimensionless quantity. The slope of the calibration curve is m = 2.24 X 10* M-'. Find the concentration detection limit with Equation 5-5. (c) Find the lower limit of quantitation with Equation 5-6.