Calculate the volume of 0.250 M H2SO4 that contains 0.250 g H2SO4. 0.250 g H2SO4 x 1 mole x 1 L = 0.0102 L 98.12 g 0.250 mole 5. 1.50 g of NaCl is dissolved in 100.0 mL of water. Calculate the concentration. 6.
(0.125)(0.1) = 0.0125 moles of solute b. Pour 50 mL of the stock solution to get the number of moles needed. 6. Exercise 6: a. (0.02) (1) = x/ 372.24 x=7.4 grams of sodium EDTA that is weighted out. b.
7. 0.50 g of CaSO4 is dissolved in 1.0 L of water. Is this solution saturated? The Ksp of CaSO4 is
29.4 atm B. 4.89 atm C. 25.1 atm D. 36.0 atm _____ 5. The vapor pressure of pure ethanol at 60 °C is 349 mm Hg. Calculate the vapor pressure at 60 °C of a solution prepared by dissolving 10.0 mol of naphthalene (nonvolatile) in 90.0 mol of ethanol. A.
Results and Discussion For the first part of the experiment (Part A), five different 100 mL volumetric flasks were each filled with 1,2,3,4 and 5 mL of iron (II) solution. Then 5 mL of YY ligand, were poured to each of the five flasks. Each flask had 5 mL of 2M sodium acetate and 4 mL of 3M NH2OH. Then the whole solution was diluted up to the 100 mL fill mark with distilled water. This was the solution that was used in order to obtain the absorption spectrum for each of the different iron (II) ligand examples different flasks.
EXPERIMENT 11 Molar Mass of a Volatile Liquid DATA Run | UNKNOW | Mass of flask and foil(g) | 90.085g | Mass of flask and foil and condensed vapor (g) | 90.640g | Temperature of boiling water (˚C ) | 98˚C | Barometric pressure (torr) | 761mm Hg | Volume of flask (mL) | 140 mL | RESULTS Mass of unknown ( condensed vapor) | 0.555g | Volume of flask (vapor)(L) | 0.14 L | Temperature of vapor(k) | 371K | Molar mass of unknown (g/mole) | 118.380 g/mole | Advance Study Assignment 1) How would each of the following errors affect the outcome of this experiment? Would it make the molar mass high or low? Give your reasoning in three sentences or less in each case. a) The hole in the aluminum foil was quite large. if you were to have a larger hole in your aluminum cap, you would be losing vapor to the room.
Mix. k. Measure out 3/4 cup of the solution from cup 3 and add it to cup 4. Mix. l. What are the relative salt concentrations of cups 1–4? Example: Cup 2 is made up of half stock solution and half tap water, which is a 50 percent relative salt concentration.
0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.
5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution. Using our average
Percent Yield Calculations 1) Balance this equation and state which of the six types of reaction is taking place: ____ Mg + ____ HNO3 ( ____ Mg(NO3)2 + ____ H2 Type of reaction: __________________________ 2) If I start this reaction with 40 grams of magnesium and an excess of nitric acid, how many grams of hydrogen gas will I produce? 3) If 1.7 grams of hydrogen is actually produced, what was my percent yield of hydrogen? 4) Balance this equation and state what type of reaction is taking place: ____ NaHCO3 ( ____ NaOH + ____ CO2 Type of reaction: __________________________ 5) If 25 grams of carbon dioxide gas is produced in this reaction, how many grams of sodium hydroxide should be produced? 6) If 50 grams of sodium hydroxide are actually produced, what was my percent yield?