1. chemistry

    A student performed the freezing point depression experiment according to directions but mistakenly recorded the mass of lauric acid as 8.300g instead of8.003g. The calculated molar mass of the unknown acid will be GREATER/LESS than true value b/c mass of lauric acid appears ...
  2. chemistry

    (A) A student performed the freezing point depression experiment according to directions but mistakenly recorded the mass of lauric acid as 8.300 g instead of 8.003 g. The calculated molar mass of the unknown acid will be ___________ than the true value, because the mass of ...
  3. chemistry

    A) A student performed the freezing point depression experiment according to directions but mistakenly recorded the mass of lauric acid as 8.300 g instead of 8.003 g. The calculated molar mass of the unknown acid will be ___________ than the true value, because the mass of ...
  4. chemistry

    1) The color of bromocresol green indicator is??? when pH < 4, green when pH 4- 5.3 and ???? in pH > 5.3. 2)(A) A student performed the freezing point depression experiment according to directions but mistakenly recorded the mass of lauric acid as 8.300 g instead of 8....
  5. lauric acid

    what happens to the calculated molar has if the lauric acid is spilled before determining the freezing point?
  6. Chemistry HELP!

    a) How would a too low molarity of NaOH affect the calculated molar mass of the unknown acid? b) How would it affect the calculated Ka of the unknown acid? If mistakenly calibrated pH meter using pH6 buffer instead of pH7 buffer, all ph meter readings were too low.... how does...
  7. Chemistry HELP!

    a) How would a too low molarity of NaOH affect the calculated molar mass of the unknown acid? b) How would it affect the calculated Ka of the unknown acid? If mistakenly calibrated pH meter using pH6 buffer instead of pH7 buffer, all ph meter readings were too low.... how does...
  8. ap chemistry

    what would have happened to your calculated molar mass if some lauric acid was spiled before finding the freezing point?
  9. chemistry

    The procedure described in this experiment was used to determine the molar mass of unknown liquid (non-electrolyte). The solution was made by mixing 0.961 g of the unknown with 100.0 g of water. The freezing point depression of the solution was −3.7°C. Calculate the ...
  10. Chemistry

    Since the unkown acid is monoprotic, this also equals the number of moles of acid to use. A typical molar mass for an unknown acid in this experiment is 380 g/mol. Using this molar mass, calculate the mass (in grams) of unknown acid you should use per titration. 1.Suppose that...
  11. Chemistry

    We will calculate the amount of acid to use in each titration. Assume that you are using 0.0512 M NaOH(aq). A good volume of NaOH(aq) to use per titration is 15 mL. From this molarity and volume, the moles of NaOH can be calculated. Since the unknown acid is monoprotic, this ...
  12. Chemistry

    Experiment questions: Experiment 1 results mass of beaker:116.944 g mass of dodecanoic:11.629 g hot water: 175 C warm water: 10 C Average Fressing pt of Dodecanoic = 42.7 Experiment 2 results: Unknown C solution mass: 2.091g Hot water: 175C warm water: 10C average FP of UK C ...
  13. Chemistry

    Calculate the expected freezing point of a solution of 0.579 g pelargonic acid in 4.225 g of lauric acid, if the melting point of pure lauric acid was measured to be 43.20*C. (MM pelargonic acid = 158.2 g/mol; MM laruic acid = 200.32 g/mol). Kf for lauric acid is 3.90*C/Mc; ...
  14. College Chemistry

    We will calculate the amount of acid to use in each titration. Assume that you are using 0.0512 M NaOH(aq). A good volume of NaOH(aq) to use per titration is 15mL. From this molarity and volume, the moles of NaOH can be calculated. Since the unknown acid is monoprotic, this ...
  15. Chem

    We will calculate the amount of acid to use in each titration. Assume that you are using 0.0512 M NaOH(aq). A good volume of NaOH(aq) to use per titration is 15 mL. From this molarity and volume, the moles of NaOH can be calculated. Since the unknown acid is monoprotic, this ...
  16. Chemistry

    A student completes a titration of an unknown diprotic acid. In this experiment, 0.79 g of the acid is dissolved in 250.0 mL of water. It requires 13.48 mL of 1.0 M NaOH to reach the second equivalence point. What is the molar mass of the acid?
  17. ap chemistry

    what would have happened to the calculated molar mass if some benzoic acid was spilled before adding it to the lauric acid? also, what would have happened to the calculated molar mass if the thermometer used actually read 1.4C too high I NEED HELP PLEASE THANKS
  18. Chemistry-Dr.Bob222

    A student completes a titration of an unknown diprotic acid. In this experiment, 0.79 g of the acid is dissolved in 250.0 mL of water. It requires 13.48 mL of 1.0 M NaOH to reach the second equivalence point. What is the molar mass of the acid?
  19. chemistry

    A freezing point depression experiment was conducted using cyclohexane as the solvent. The freezing point of pure cyclohexane is 6.60°C and the freezing point depression constant is 20.00°C/m. The freezing point of a solution containing 0.135 g of an unknown nonelectrolyte ...
  20. CHEMISTRY- PLEASE HELP:)!

    Pure lauric acid is melted and the freezing point is determined to be 43 degrees celsius. A solution is made by dissolving .50 grams of paradicholrobenzene into 3.00 grams of lauric acid. Determine the molality of this solution. I got 1.1m. which is right. But then it says the...
  21. Chemistry

    The molal freezing point constant(Kf) for acetic acid is 3.90 C/m. When 31.5 grams of unknown solute is dissolved in 650 grams of acid the freezing point is lowered by 0.79 what is the molar mass
  22. Chemistry

    An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% Carbon and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20 ...
  23. Chem

    Using the standardized NaOH solution, the student weighs out 0.2550g of a solid unknown acid and finds that 28.50mL of the NaOH solution is required to reach an end point with the acid. Calculate the molar mass of the unknown acid, assuming the acid is diprotic.
  24. Chem

    Using the standardized NaOH solution, the student weighs out 0.2550g of a solid unknown acid and finds that 28.50mL of the NaOH solution is required to reach an end point with the acid. Calculate the molar mass of the unknown acid, assuming the acid is diprotic.
  25. Chemistry

    Determine the molar mass of the unknown acid.? A 0.167-g sample of an unknown acid requires 27.8 mL of 0.100 M NaOH to titrate to the equivalence point. Elemental analysis of the acid gives the following percentages by mass: 40.00% C; 6.71% H; 53.29% O.
  26. Chemistry

    Determine the molar mass of the unknown acid.? A 0.167-g sample of an unknown acid requires 27.8 mL of 0.100 M NaOH to titrate to the equivalence point. Elemental analysis of the acid gives the following percentages by mass: 40.00% C; 6.71% H; 53.29% O.
  27. Chemistry

    Find the Molecular weight of unknown acid #5... Mass of acid #5 = 1.200g Volume (acid was mixed in) = 100ml (distilled water) Conectration of NaOH = 0.989 M Aliquot of acid titrated with NaOH = 25ml = 0.025 L Average volume of NaOH from titration = 9.3ml = 0.0093L Molar ratio ...
  28. Chemistry II

    An unknown solute (a nonelectrolyte) was obtained and 5.37 g was weighed out. After dissolving the solute in water, the mass of the solution was 26.58 g. From the experimental results, the freezing point depression was found to be 3.6oC. If the freezing point depression ...
  29. Chemistry

    A 0.120 g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1218 M NaOH. The equivalence point is reached after adding 12.4 mL of base. What is the molar mass of the unknown acid?
  30. Chemistry

    A 0.120 g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1218 M NaOH. The equivalence point is reached after adding 12.4 mL of base. What is the molar mass of the unknown acid?
  31. Chemistry

    Include an explanation, balanced chemical equation, units, significant figures and/or calculations where appropriate. Determination of the Molar Mass of an Unknown Acid Trial 1: 0.499 g unknown acid 24.07 mL NaOH Trial 2: 0.501 g unknown acid 23.71 mL NaOH Trial 3: 0.500 g ...
  32. Chemistry

    A student titrated a solution containing 3.7066 grams of an unknown triprotic acid to the end point using 28.94 milliliters of 0.3021 M KOH. What is the molar mass of the unknown acid? And write a balanced equation for the reaction.
  33. chemistry h/w

    Hi, this is my first lab in 2nd semester of chemistry, and im not sure about my answers in the assignment can someone give it a quick look to check them? this is my phase diagram: tinypic. com/view.php?pic=1605yev&s=7 p.s. ignore the lone red dot its supposed to be 38C.. there...
  34. Chemistry h/w

    Hi, this is my first lab in 2nd semester of chemistry, and im not sure about my answers in the assignment can someone give it a quick look to check them? this is my phase diagram: tinypic. com/view.php?pic=1605yev&s=7 p.s. ignore the lone red dot its supposed to be 38C.. there...
  35. Chemistry (College) URGENT

    1.6g of an unknown monoprotic acid (HA) required 5.79 mL of a 0.35 M NaOH solution to reach the equivalence point. Determine the molar mass of the acid and given that the pH at the half way point to the equivalent point is 3.86. Calculate the Ka of the unknown acid.
  36. Chemistry

    I got think I got the following questions almost figured, but I just want someone to make sure that I am correct. 1) The whole experiment involves using the ideal gas law to predict the molar mass of an unknown liquid. The formula is molar mass of liquid = gR(0.0821)T(...
  37. Chemistry

    When 0.500 g of an unknown compound was dissolved in 15.0 g benzene, the freezing point depression was determined to be 0320C . The molar mass of the unknown compound is ? g/mol. (The freezing point depression constant for benzene is 512C kgmol .
  38. chem

    A solution prepared by dissolving 3.00 grams of ascorbic acid (vitamin C, C6H8O6), in 50.0 grams of acetic acid has a freezing point that is depressed by T = 1.33 oC below that of pure acetic acid. What is the value of the molar freezing point-depression constant for acetic ...
  39. chemistry

    When 0.500 g of an unknown compound was dissolved in 15.0 g benzene, the freezing point depression was determined to be 0.320 C. The molar mass of the unknown compound is ____ (The freezing point depression constant for benzene is 5.12 C kg/mol. )
  40. URGENT -- Chemistry

    When 0.500 g of an unknown compound was dissolved in 15.0 g benzene, the freezing point depression was determined to be 0320C. The molar mass of the unknown compound is ?? g/mol. (The freezing point depression constant for benzene is 5.12C kgmol. I really don't know where to ...
  41. Chemistry

    I wanted to calculate the average equivalent weight of an unknown acid for an acid-base titration experiment. The mass of the unknown that I had to obtain is 0.2 g and the equivalents of the acid is .001046 L x N. But is this equation right? Does the equivalent mass equal to ...
  42. Chemistry - Colligative properties/freezing point

    I need help on my lab. I think I have all the data and steps right, but my results aren't making sense: Feezing point of water: -0.58 C Freezing point of Unknown: -0.93 C Freezing point depressions: 0.34 C Mass of unknown: 1.045 g, mass of solvent (water): 19 ml, Water=1.86 C/...
  43. chemistry

    1. Find the molality of the solution prepared by dissolving 0.238g toluene, C7H8, in 15.8g cyclohexane 2. A pure sample of the solvent phenol has a freezing point of 40.85 degrees C. A 0.414 molal solution of isopropyl alcohol was observed to have a freezing point of 38.02 ...
  44. chemistry

    The following errors occurred when the experiment was carried out. How would each affect the calculated molar mass of the solute(too high, too low, no effect)? Explain. a. The thermometer used actually read 1.4 centigrade too high. b. some of the solvent was spilled before the...
  45. chemistry... please help!

    Pure glacial acetic acid HC2H3O2 has a freezing point of 16.62 degrees C. Its freezing point depression constant is Kf=3.57 Cm^-1. A solution was made by taking 9.755 g of an unknown non-electrolyte and dissolving it in 90.50 g of glacial acetic acid. The measured freezing ...
  46. CHEM- FP

    camphor melts at 179.8 degrees, and it has a particularly high freezing point depression constant.(kf=40 degrees/m). when 0.186g of an organic substance of an unknown molar mass is dissolved in 22.01g of liquid camphor, the freezing point of the mixture is found to be 176.7 ...
  47. Chemistry

    What effect on the molar mass of the unknown acid would each of the following have-that is-would it make the calculated molar mass high, low, or would it have no effect? 1. Adding 5 drops of phenolphthalein instead of 1 drop. 2. Using a standard NaOH solution that has been ...
  48. chem

    if the freezing point of the solution is recorded 0.2 degrees C lower than the actual freezing point, will the molar mass determination for the unknown solid be too high or too low?
  49. Chemistry

    Suppose that 0.483 g of an unknown monoprotic weak acid, HA, is dissolved in water. Titration of the solution with 0.250 M NaOH(aq) required 42.0 mL to reach the stoichiometric point. After the addition of 21.0 mL, the pH of the solution was found to be 3.75. (a) What is the ...
  50. Chemistry

    A 0.125-g sample of monoprotic acid of unknown molar mass is dissolved in water and titrated with 0.1003 M NaOH. The endpoint is reached after adding 20.77 mL of base. What is the molar mass of the unknown acid?
  51. chemisry

    A 0.125 gram sample if a monoprotic acid of unknown molar mass is dissloved in water and titrated with 0.1003 MNaOH. The endpoint is reached after adding 20.77 ml of base. What is the molar mass of the unknown acid?
  52. APChem

    If an air bubble passes unnoticed through the tip of a buret during a titration (where a weak nonvolatile acid HA is the analyte) , the calculated molar mass of HA would be too high, too low, or not affected? I thought it would be too high, since an air bubble in the buret ...
  53. Chemistry

    Titration of an Unknown Acid... What effect on the molar mass of the unknown acid would each of the following have... Would it make the calculated molar mass high, low, or would it have no effect? 1. Adding 5 drops of phenolphthalein instead of 1 drop... Answer: No effect. 2. ...
  54. I am so lost

    The freezing point of pure cyclohexane is 6.60°C and the freezing point depression constant is 20.00°C/m. The freezing point of a solution containing 0.161 g of an unknown nonelectrolyte solute and 5.818 g of cyclohexane was 3.98°C. What is the molar mass of the unknown ...
  55. I am so lost

    The freezing point of pure cyclohexane is 6.60°C and the freezing point depression constant is 20.00°C/m. The freezing point of a solution containing 0.161 g of an unknown nonelectrolyte solute and 5.818 g of cyclohexane was 3.98°C. What is the molar mass of the unknown ...
  56. Chemistry

    A .288 g sample of an unknown monoprotic organic acid is dissolved in water and titrated with a .115 M sodium hydroxide solution. After the addition of 17.54 mL of base, a pH of 4.92 is recorded. The equivalence point is reached when a total of 33.83 mL of NaOH is added. What ...
  57. chemistry

    Suppose that a student performing this experiment recorded the concentration of NaOH solution as 0.513 M, instead of the correct value of 0.531 M. Assuming there were no other errors, would each of the following experimental results be high, low, or unaffected by the error? ...
  58. Chem

    Suppose that 16.07 mL of 0.0512 M NaOH were required to titrate a sample of unknown acid. How many moles of NaOH were used? Assuming that the unknown acid sample in question 1 had a mass of 0.177 g, what is the molar mass of the unknown acid?
  59. Chemistry

    Calculate amount (in grams) of unknown acid used for titration. You want 30 mL of titrant to get the equivalence point. Assume that the base is 0.05 M and the unknown acid molar mass is around 400 g/mol.
  60. Chemistry

    A 0.5224g sample of an unknown monoprotic acid was titrated with 0.0998M of NaOH. The equivalence point of the titration occurs at 23.82 mL. Determine the molar mass of the unknown acid.
  61. Chemistry

    A 0.120 g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1218 M NaOH. The equivalence point is reached after adding 12.4 mL of base. What is the molar mass of the unknown substance?
  62. Chemistry

    I there, this is my lab worksheet, I am having trouble filling it out. Please help me with it. Expt #1- Molecular Weight of Unknown Acid Unknown Acid: #3 Mass of Unknown solid acid transferred:1.0g Volume of volumetric flask: 100.00 mL Concentration of NaOH: 0.0989 M Aliqot of...
  63. Chemistry

    Item 1 Maleic acid is a carbon-hydrogen-oxygen compound used in dyeing and finishing fabrics and as a preservative of oils and fats. •In a combustion analysis, a 1.054-g sample of maleic acid yields 1.599 g of CO2 and 0.327 g of H2O. •When 0.609 g of maleic acid is ...
  64. Chemistry

    A .682 g sample of an unknown weak monoprotic acid, HA, was dissolved in sufficient water to make 50mL of solution and was titrated with a .135 M NaOH solution. After the addition of 10.6mL of base, a pH of 5.65 was recorded. The equivalence point was reached after the ...
  65. Chem

    A sample contains an unknown amount of tartaric acid, H2C4H4O6. If 0.3888 g of the sample requires 37.74 mL of 0.1000 M NaOH to neutralize the H2C4H4O6 completely, what is the percentage of H2C4H4O6 in the sample? The molar mass of H2C4H4O6 is 150.09 g/mol. The balanced ...
  66. chem

    A 0.167-g sample of an unknown compound contains 0.00278 mol of the compound. Elemental analysis of the acid gives the following percentages by mass: 40.00% C; 6.71% H; 53.29% O. 1.)Determine the molecular formula of this acid. 2.)Determine the molar mass of the unknown acid.
  67. Chemistry

    Five (5.00) grams of glucose, C6H12)6, is dissolved in 500.0 grams of acetic acid. What is the new freezing point and boiling point for the solution? Kf acetic acid = 3.90, Kb acetic acid = 3.07 (normal freezing point for acetic acid = 16.60 dg C, boiling point = 118.5 dg C) f...
  68. Chemistry

    A 0.2800 g sample of an unknown acid requires 28.22 mL of 0.1199 M NaOH for neutralization to a phenolphthalein point. a.) How many moles of OH- are used? b.) How many moles of H+ are found in the acid? c.) What is the calculated equivalent molar mass of the unknown acid? If I...
  69. Chemistry

    Assuming that the unknown acid sample in question 1 had a mass of 0.177 g, what is the molar mass of the unknown acid?
  70. Chemistry

    In a experiment to determine the molecular weight and the Ka for ascorbic acid (vit. c.) a student dissolved 1.3713g of the monoprotic acid in water to make 50 mL of solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 mL of the...
  71. Chemisty - Colligative properties/freezing point

    QUESTION: How do you determine the molar mass of the unknown, nonelectrolyte compound for each trial? This is for a lab. My data is: freezing point of solvent (water): -0.58 freezing point of unknown: -0.93 Mass of solvent (water): 19 ml Mass of unknown: 1.045g I know the ...
  72. Chemistry

    A 0.2800 g sample of an unknown acid requires 28.22 ml of 0.1199 M NaOH for neutralization to a phenolphthalein end point. a.) How many moles of OH- are used? b.) How many moles of H+ are found in the acid? c.) What is the calculated equivalent molar mass of the unknown acid? ...
  73. chemistry, freezing point depression

    Hey guys I did a lab and I have to answer the questions relating to the lab. I did question 1 & 2, can someone check if they are correct, and question 3 I don't know how to do it. 1. Calculation to Determine the molecular weight of unknown substance Mass of unknown used: 2.0 g...
  74. Science (Chemistry) (Measurement Uncertainty)

    Three students are asked to determine the mass of a coon. They each measure different values for the mass of the coin using a balance. Student A recorded a mass of 63.08g, student B recorded a mass of 63.022g and student C recorded a mass of 63.0328g. If the balance is ...
  75. Chem

    If .00840 mol is the answer to this question (Suppose that 16.41 mL of 0.0512 M NaOH were required to titrate a sample of unknown acid. How many moles of NaOH were used? ) how would I answer this: Assuming that the unknown acid sample in question 1 had a mass of 0.177 g, what ...
  76. Chemistry

    Expt #1- Molecular Weight of Unknown Acid Unknown Acid: #2 Mass of Unknown solid acid transferred:0.414g Volume of volumetric flask: 100.00 mL Concentration of NaOH: 0.0989 M Aliqot of acid titrated with NaOH: 25.00 mL Average volume of Naoh from titration: 13.9 mL Here's ...
  77. CHemistry....Please help

    1.SOLUBILITY: At 25C, the molar solubility of calcium phosphate in water is 1.1*10^{-7}M. Calculate the solubility in grams per liter. I m not sure how to convert M to g/L MOLAR MASS OF COLLIGATIVE PROPERTIES: 2.tert-Butyl alcohol is a solvent with a Kf of 9.10 C/m and a ...
  78. chemistry

    .256g unknown acid is titrated with .12M NaOH. It requires 34.8ml of base to reach the equivalence point. What is the molar mass of the acid?
  79. chemistry

    a solution is prepared by mixing 2.17g of an unknown non-electrolyte with 225.0g of chloroform. The freezing point of the resulting solution is -64.2 C . The freezing point of pure chloroform is -63.5 C and its kf= 4.68C m^-1 . What is the molecular mass of the unknown? i know...
  80. Chemistry

    If 20mL of a 0.100 M NaOH solution are required to titrate 0.25 grams of an unknown acid, what is the equivalent mass of the acid? If the acid actually contains 3 moles of H+ per mole of acid what is the molecular mass of the acid?
  81. ap chem

    Find the molecular mass of a solute by freezing point depression. Solvent: para-dichlorobenzene Freezing point of pure solvent: 53.02C Mass of unknown substance: 2.04g Freezingpointdepressin constant:7.1c/m Mass of para-dichlo...: 24.80g Freexing point solution: 50.78 C Can ...
  82. chemistry

    Suppose that 0.483 g of an unknown monoprotic weak acid, HA, is dissolved in water. Titration of the solution with 0.250 M NaOH(aq) required 42.0 mL to reach the stoichiometric point. After the addition of 21.0 mL, the pH of the solution was found to be 3.75. (a) What is the ...
  83. College Chemistry

    My professor gave us two homework questions that are a bit confusing, can you help me to get started on them. 1) You have 6g of water (Kf=1.86 C kg/mol). You add and dissolve 2.1 g of a different unknown substance. The freezing point of the solution is -0.3 deg C. What is the ...
  84. Chemistry

    A mass of 0.4113 g of an unknown acid, HA, is titrated with NaOH. If the acid reacts with 28.10 mL of 0.1055 M NaOH, what is the molar mass of the acid?
  85. chemistry

    a 3.54 grams solid sample of an unknown monoprotic acid was dissolved in distilled water to produce a 47.0 mL solution at 25 degrees. This solution was then titrated with 0.2 M NaOH. The equivalence point was reached when 35.72 mL of 0.2 M NaOH was delivered. A. find the ...
  86. chemistry help

    a 3.54 grams solid sample of an unknown monoprotic acid was dissolved in distilled water to produce a 47.0 mL solution at 25 degrees. This solution was then titrated with 0.2 M NaOH. The equivalence point was reached when 35.72 mL of 0.2 M NaOH was delivered. A. find the ...
  87. chemistry

    a student determines the molar mass of methanol by the method used in this experiment. she found that the equilibrium temperature of a mixture of ice and pure water was .4 degrees Celsius on her thermometer. when she added 10g of her sample to the mixture, the temperature, ...
  88. Chemistry

    We did a titration lab, and one of the post-lab questions says: "the manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Use your results and a density of 1g/mL to determine if this claim is true or false. So the ...
  89. Chemistry

    We did a titration lab, and one of the post-lab questions says: "the manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Use your results and a density of 1g/mL to determine if this claim is true or false. So the ...
  90. Chemistry 1200

    Freezing-point depression can be used to determine the molecular mass of a compound. Suppose that 1.28 g of an unknown molecule were added to 19.9 g of water and the freezing point of the solution determined. If the new freezing point of water were found to be -1.50°C, what ...
  91. Chem 1200

    Freezing-point depression can be used to determine the molecular mass of a compound. Suppose that 1.28 g of an unknown molecule were added to 19.9 g of water and the freezing point of the solution determined. If the new freezing point of water were found to be -1.50°C, what ...
  92. Chemistry

    Unknown Acid: #2 Mass of Unknown solid acid transferred:0.414g Volume of volumetric flask: 100.00 mL Concentration of NaOH: 0.0989 M Aliqot of acid titrated with NaOH: 25.00 mL Average volume of Naoh from titration: 13.9 mL Here's where I need help: No of moles NaOH used: ? ...
  93. Chemistry

    The molar mass of a non-dissociated, non-volatile compound is to be determined by dissolving 5.00 g of the compound into 50.0 g of benzene. The freezing point of benzene was lowered by 2.5 degrees C. The freezing point depression constant (Kf) for benzene is 5.0 degree C/molal...
  94. Chemistry

    10.0 g of an unknown acid is titrated to the equivalent point using 65.0 ml of 0.150M KCH. Assuming the acid is monoprotic ( like HCL or HF ) , what is the molecular mass of the acid? Is it possible to find the mols of the acid? (Balanced equation & solution please!)
  95. Chemistry

    10.0 g of an unknown acid is titrated to the equivalent point using 65.0 ml of 0.150M KCH. Assuming the acid is monoprotic ( like HCL or HF ) , what is the molecular mass of the acid? Is it possible to find the mols of the acid? (Balanced equation & solution please!)
  96. Science

    It is found that 24.68 mL of 0.1165 NaOH is needed to titrate 0.2931 g of an unknown acid to the phenolphthalein end point. What is the equivalent mass of the acid? equivalent mass = grams acid/moles H+
  97. CHEMISTRY II

    How would the calculated value of the molar mass of an unknown acid be affected (higher, lower, or no change) if the following occurs? (a) The pH meter was incorrectly calibrated to read lower than the actual pH. (b) During the titration, several drops of NaOH missed the ...
  98. Chemistry lab

    A 1.340gram sample of an unknown acid, H3A, was placed in a 250.0 mL volumetric flask and diluted to volume with water. A 45.35 mL sample of this acid solution was titrated with 37.77 mL of a 0.1006M NaOH solution. A) Using one set - up dimensional analysis B) Calculate the ...
  99. Chemistry

    After you preform your experiment, you determine that the Kf value for naphthalene is 6.89 °C/m . You are using 10g of naphthalene and added 1.0 g of your unknown. The the freezing point of the solvent decreased by 4.42 °C when the unknown was added. Knowing this information...
  100. chemistry

    After you preform your experiment, you determine that the Kf value for naphthalene is 6.9 . You are using 10g of naphthalene and added 1.0 g of your unknown. The the freezing point of the solvent decreased by 4.42 when the unknown was added. Knowing this information, determine...
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  89. 88
  90. 89
  91. 90
  92. 91
  93. 92
  94. 93
  95. 94
  96. 95
  97. 96
  98. 97
  99. 98
  100. 99
  101. 100