What is the concentration of NaOH (sodium hydroxide), in ppm (m/m), in a solution prepared by dissolving 32 mg of NaOH in 5.8 kg of water? Report your result in decimal notation and to the proper number of significant figures. All numbers are measured. a) 0.18 ppmb) 1.8 ppmc) 0.55 ppmd) 5.5 ppme) no correct response
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What is the concentration of NaOH (sodium hydroxide), in ppm (m/m), in a solution prepared by dissolving 32 mg of NaOH in 5.8 kg of water?
Report your result in decimal notation and to the proper number of significant figures. All numbers are measured.
a) 0.18 ppm
b) 1.8 ppm
c) 0.55 ppm
d) 5.5 ppm
e) no correct response
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- What is the percent by volume (% v/v) of ethanol in a solution prepared by adding enough water to 75.0 mL of ethanol to make 350.0 mL of ethanol solution? Report your result in decimal notation and to the proper number of significant figures. All numbers are measured. a) 7.50% b) 21.4% c) 75.0% d) 27.3% e) no correct response3:36 1 ul LTE O Send a chat What volume in of a 0.32 M Mg(NO3)2 solution contains 45 L g of Mg(NO3)2 ? STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 210.34 6.022 × 10" 45 0.30 140 0.32 1 10 148.33 0.95 86.32 0.097A chemistry student weighs out 0.0312 g of sulfurous acid (H, SO,, a diprotic acid, into a 250. mL volumetric flask and dilutes to the mark with distilled w He plans to titrate the acid with 0.1000M NAOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Round your answer to 3 significant digits. mL x10 Explanation Check O 2022 McGraw Hill LLC. AII Rights Reserved. Terms of Use | Privacy Center | Ac 80°F Cloudy DELL F7 F8 F9 F10 F11 F12 PrtScr Insert Delete PgL F1 F2 F3 F4 F5 F6 Fn @ %2# & Backspace
- Question 16 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz 354.9 mL The following data were obtained: Peak Area Benzoate Caffeine Aspartame Std 1 2.30 3.54 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…What is the pK* in the average human blood serum which contains 18 mg of K+ per 100 mL? O 0.74 2.34 O 3.64 O 4.14 no correct response * Previous NextNEED ANSWER PLS FAST 50 g of NaCl has been dissolved in water to come up with a 120 mL of solution. The saline solution has a density of 1.3 g/mL. What is the concentration of the solution in terms of M? Responses A 6.27 M B 9. 17 M C 7.17 M D 0.66 M
- Question 14 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Caffeine Peak Area Benzoate 3.54 Aspartame Std 1 2.30 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…Page of 2 Practice filling in the chart below. When you are finished, check your answers with the key at the front. Then do the retest on the back. pH pOH [H*] (OH) 1. 6.51 2. 10.32 3. 9.32 X 103 4. 4.71 X 10-8Question 15 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 µL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Peak Area Benzoate Caffeine Aspartame Std 1 2.30 3.54 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…
- Question 17 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Caffeine Peak Area Benzoate 3.54 Aspartame Std 1 2.30 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…Question 12 Caffeine, benzoate, and aspartame content of mountain dew soda was determined using reverse phase HPLC. The concentration of the standard used are as follows, caffeine: 0.70 mg/mL, benzoate: 1.6 mg/mL, and aspartame 5.0 mg/mL. The following volume of the standards were taken to a 50.0 mL volumetric flask and diluted to the mark to construct a calibration curve: Caffeine Benzoate Aspartame 1 mL 1 mL 1 mL 2 mL 2 mL 2 mL 3 mL 3 mL 3 mL 4 mL 4 mL 4 mL 5 mL 5 mL 5 mL The 12-oz can mountain dew sample were treated as follows: The 12- oz can were has been left out overnight to get rid the carbonation then, 2 mL were drawn into a plastic syringe, filtered, and placed into a vial. An equal amount of deionized water was added. A 100 μL sample were injected into a sample loop using same parameters with the standards. Assume 12 oz = 354.9 mL The following data were obtained: Peak Area Benzoate Caffeine Aspartame Std 1 2.30 3.54 3.33 Std 2 4.10 5.59 5.83 Std 3 7.37 9.98 9.63 Std 4 9.56…Trial 1 Calculations Weight of Antacid 1.541 g Concentration of HCl used 0.500 M Vol HCl added (mL) 49.67 mL Concentration of NaOH 0.453 M Total moles of HCl added 9.02 Moles HCl = Molarity x volume in liters Initial buret reading (NaOH) 25.65 mL Final buret reading (NaOH) 49.72 mL Vol NaOH added (mL) Final -Initial buret readings (NaOH) Moles of NaOH used Moles NaOH used = Molarity x volume in L Moles of HCl neutralized by NaOH 10.90x103 moles Moles of HCl neutralized by NaOH = moles NaOH (1mole acid/1 mole base) Moles of HCl neutralized by antacid Moles of HCl neutralized by the antacid = total moles HCl added – moles HCl neutralized by NaOH Moles of HCl consumed per gram of antacid Moles of HCl consumed per gram of antacid= moles HCl neutralized by antacid/ grams of antacid subtract the moles of HCl nuetralized by the NaOh from the total moles of HCl added…