#1: Rubbing alcohol is sold over the counter as a topic (skin) antiseptic. Rubbing alcohol is typically 70% (v/v) iso-propyl alcohol, How much of this solution can you prepare with 500 mL of pure C3H;OH? Again, most concentration problems will have two (2) quantities. Percent by volume (% v/v) is not a good starting point because it is n factor). So, look at the other quantity as your dimensional analysis start point. Write the volume quantity, units, and "of what", over

#1: Rubbing alcohol is sold over the counter as a topic (skin) antiseptic. Rubbing alcohol is typically 70% (v/v) iso-propyl alcohol, How much of this solution can you prepare with 500 mL of pure C3H;OH? Again, most concentration problems will have two (2) quantities. Percent by volume (% v/v) is not a good starting point because it is n factor). So, look at the other quantity as your dimensional analysis start point. Write the volume quantity, units, and "of what", over

Introduction to General, Organic and Biochemistry

11th Edition

ISBN:9781285869759

Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Chapter6: Solutions And Colloids

Section: Chapter Questions

Problem 6.34P

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6-25 A small amount of solid is added to a separatory funnel containing layers of diethyl ether and water. After shaking the separatory funnel, in which layer will we find each of the following solids? (a) NaCI (b) Camphor (C10H16O) (c) KOH
6-71 A 4 M acetic acid (CH3COOH) solution lowers the freezing point by-8°C; a 4 M KF solution yields a-15°C freezing-point depression. What can account for this difference?
5. A water-insoluble compound is discovered to be soluble in10% NaOH. Which of the following substances is most likely to be the compound? O CH;CH2-OH OH (CH), CH3 O CH3(CH2)8CH2-OH Explain your answer:
(D Page view A Read aloud V Draw A. Solubility and Salt Formation (Dissociation & Neutralization) Obtain 4 clean test tubes. Test tube 1 - Place approximately 5mL of distilled water and 1mL of acetic acid Test tube 2 - Place approximately 5mL of distilled water and ImL of oleic acid Test tube 3 - Place approximately 5mL of IM NaOH and ImL of acetic acid Test tube 4 - Place approximately 5mL of IM NAOH and ImL of oleic acid Make a prediction about each reaction in the table below. Neutralization with NaOH Dissociation in Water Compound (draw the structure AND name the salt formed) (yes/no) Acetic acid Yes CH;COOH Oleic acid No CH:(CH2)-CH-CH(CH2) CO2H hp
The following compounds are available as 0.10 M aqueous solutions: pyridine triethylamine m-nitroaniline NaOH phenol HCIO HCIO4 (рКь = 8.82) (pKb = 3.25) (pKb = 11.53) (K₂ = 1.1 × 10-10) (K₂ = 2.9 × 10-8) Which two solutions could be combined to prepare a pH 6.7 buffer? HCIO and NaOH phenol and NaOH m-nitroaniline and HCIO4 pyridine and HCIO4
answers are: (a) 5.34 g camphor and 8.01 g castor oil(b) 8.39 mL castor oil(c) 3.12% w/v lactic acid, 3.91% w/v salicylic acid, and 1.56% w/v trichloroacetic acid i want to know the full solutions. thank you!
Answer is provided. All I need is solution. Thank you From the following formula, calculate the quantities required to make 5 lb of hydrophilic ointment. Methylparaben 0.25 g Propylparaben 0.15 g Sodium lauryl sulfate 10 g Propylene glycol 120 g Stearyl alcohol 250 g White petrolatum 250 g Purified water, ad 1000 g Answer: 0.568 g methylparaben 0.341 g propylparaben 22.7 g sodium lauryl sulfate 272.4 g propylene glycol 567.5 g stearyl alcohol 567.5 g white petrolatum ad 2270 g or 5 lb purified water
Materials Needed solid I2 solid CUSO4-5H20 food dye solid (NH4)2SO4 heavy metals waste container halogenated waste container non-halogenated waste container semi-micro test tubes and rack regular test tubes and rack squash pipettes acetone cyclohexane propan-2-ol Method Part A: Solubility of ionic and molecular solids 1. Place a small amount (about the size of 1 grain of rice, see picture) of copper sulfate into each of three DRY semi-micro test tubes. Add 20 drops of water to the first test tube and gently flick the test tube with your finger to ensure mixing. 2. Repeat step 1 using acetone in place of water as the solvent in the second test tube. 3. Repeat step 2 replacing acetone with cyclohexane in the third test tube. Hold the test tubes against a white background to compare the solubility of copper sulfate in the three solvents and record your results. Discard these mixtures into the heavy metals waste container in the fume cupboard. Once these test tubes have been emptied you…
Assume you have 0.1 M NaOH Trial 1: 22.5 mL NaOH Trial 2: 25.6 mL NaOH Trial 3: 24.4 mL NaOH
Question 3 3. Which of the following statement/s is/are correct? 1. 6.00 N H2CO3 is equivalent to 3.00M H2CO3. 1I.5.29 % (w/v) H2SO4 is equivalent to 0.539 N H2SO4 III. 1.15 N Mg(OH)2 is equivalent to 16.3 g Mg(OH), in 500 mL solution. (MH2504 = 98.086 g/mol; MH2C03 = 62.026 g/mol; MME(OH)2 = 58.326 g/mol) O l and II O l only O Il and III OI and II
Procedure A: Molarity (M=g/MM/L) Step 1. Place one-eighth of a teaspoon (this weighs approximately 0.5 grams) of NaCl (table salt) in the graduated cylinder provided. Then add distilled water to that graduated cylinder until the level of the water reaches 45.00 mL. Mix thoroughly about 1 minute. Pour solution into a somewhat clear cup. Remember what this looks like for Procedure B. Questions: Is this a saturated solution? Why or why not? Take a picture and insert a picture (compress your image) of your solution here. Save this solution for Step 2. Calculations: Determine the Molarity of this solution using the formula M = g/MM/L. You know the grams (one-eighth of a teaspoon weighs approximately 0.5 grams), look up the Molar Mass of NaCl, and convert your mL to L to solve the problem.) Show your calculations for Molarity here. C % 31 D < hp & * ✔
Rank a series of molecules by expected solubility in water based on polarity and hydrogen bonding. Some slightly soluble compounds are included in this exercise. Rank the organic compounds from most soluble to least soluble. To rank items as equivalent, overlap them. View Available Hint(s) Reset Help Η Η Η Η H H H H ннно Н-С-С — С —С— ОН Н-С-С-С —С -Н Н-С—С—С — С -ОН Η Η Η Η Η Η Η Η Η Η Η нн Η Η Н-С—С— О—С—С — Н нн нн Most soluble Least soluble