To do this, first take the number of drops used to achieve the monolayer (1 drop) and convert it to mL using the calibrated number of drops per mL. Then multiply the number of grams of sodium stearate per milliliter of solution. Finally, convert to moles through the molar mass of sodium stearate. HINT: The molar mass of sodium stearate is 296.5 g/mole. Answer = 1.06*10^7 moles/top layer 5.
: Se, calculated molar mass 78.93 g] 4) A 0.3528 gram mixture of H2SO4 (molar mass 98 grams) and H3PO4 (molar mass 98 grams) was mixed with a little water and titrated with 9.70 mL of 1.000 M NaOH: 3NaOH(aq) + H3PO4(aq) →3H2O(l) + Na3PO4(aq) 2NaOH(aq) + H2SO4(aq) →2H2O(l) + Na2SO4(aq) What was the percent by mass H2SO4 in the original mixture? [Ans. : 30.6] 5) A 0.9030 gram sample of Z(OH)2 and 20.00 mL of 2.000 M HCl were put into a 100.0 mL volumetric flask and mixed with enough extra water to make 100.0 mL of solution. A 10.00 mL aliquot of this solution was taken and titrated with 17.64 mL of 0.05121 M NaOH. What is the identity of the element Z?
Part B: The graduated pipet’s average density at 22.3 °C was determined to be 0.9785g/mL with a percentage error of 1.89% shows the graduated pipet to be more accurate and precise. Part C: Density of an unknown NaCl solution was measured and a calibration curve used to determine the percentage of NaCl by mass in the solution. y=0.007x + 0.998 which concluded that the concentration of the sodium chloride solution was 3.14%. INTRODUCTION Anything that you can see, touch, taste or smell, occupies space and has mass, it is called matter. Matter can be a gas, a liquid,
Calculate the percent error in the molar mass value. Enter both values in the Data Table. Molar Mass Na2CO3 = 105.99 g/mol – this is the closest molar mass to what I calculated, so the unknown M2CO3 must be sodium carbonate. Percent Error: 93.08g/mol – 105.99 g/mol (100) = 12.18% error 105.99 g/mol DISCUSSION Review the procedure and list the possible sources of error that would cause either the molar mass of the unknown to be (a) too high or (b) too low. The goal of this lab was to discover the unknown group 1 metal (M) of the compound M2CO3 by dissolving the compound in water and adding a solution of calcium chloride, CaCl2 to the solution in order to precipitate the carbonate ions to reveal the molar mass of the unknown element, thus determining the identity of the unknown element.
3. Calculate the mass of baking soda,NaHCO3 , and find the percentage of baking soda in the unknown mixture. ※ Data and observations (a) Table 1: Percent yield of Na2CO3 from baking soda Percent yield of Na2CO3 from baking soda Trial 1 Trial 2 Mass of flask + NaHCO3 95.12 g 93.45g Mass of flask 94.15g 92.03g Mass of NaHCO3 0.97g 1.42g Mass of flask + Na2CO3 95.01g 93.12g Mass of Na2CO3 (actual yield) 0.86g 1.09g Mass of Na2CO3 (theoretial yield) 0.61g 0.90g % yield of Na2CO3 140% 120% (b) Table 2: Percent of baking soda in Unknown Percentage of NaHCO3 in an unknown mixture # C Trial 1 Trial 2 Mass of flask + unknown mixture #C (before heating) 95.81g 94.85g Mass of flask 94.15g 92.03g Mass of unknown mixture (before heating) 1.66g 2.82 g Mass of flask + residue (after heating) 95.20g 93.80g Mas of H2CO3 (H2O + CO2) (after heating) 0.61g 1.05g Mass of NaHCO3 in the unknown mixture 1.65g 2.84g % of NaHCO3 in the unknown mixture 99.0% 101% ※ Calculations For table 1: Theoretical yield of Na2CO3 for trial 1
Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3. The general formula of barium chloride hydrate is BaClg-nHZO, where n is the number of water molecules. Calculate the theoretical percent water for each value of n—divide the sum of the atomic masses due to the water molecules by the sum of all the atomic masses in the hydrate, and multiply the result by 100. Complete the table. | BaCl2 | BaCl2•H2O | BaCl2•2H2O | BaCl•3H2O | Sum of atomic masses (BaCl2) | 208.23 | 208.23 | 208.23 | 208.23 | Sum of atomic masses (nH2O) | 0 | 18.02 | 36.04 | 54.06 | Sum of atomic masses (hydrate) | 208.23 | 226.25 | 244.27 | 262.29 | Percent water in hydrate (theoretical) | 0% | 7.96% | 14.75% | 20.61% | In this lab we used a Balance, centigram
Hess's Law Lab Report Purpose Data and Observations Part I Heat of Solution of Solid Sodium Hydroxide Table 1 Original temperature of water (t1) 21.0ºC Final temperature of solution (t2) 26.0ºC Temperature change (t2-t1=∆t) 5.0ºC Mass of 100 mL of water 100.0g Heat evolved by reaction 2092J Mass of NaOH(s) 2.0g Moles of NaOH 0.05001250313 moles Energy per mole of NaOH 41829.52 J/mole ∆ H1 (kJ/mole) NaOH 41.83 kJ/mole Part II Heat of Reaction between Hydrochloric Acid and Sodium Hydroxide Solution Table 2 Original temperature of HCl(aq) 21.0ºC Original temperature of NaOH(aq) 21.0ºC Average original temperature (t1) 21.0ºC Final temperature of solution (t2) 34.5ºC Temperature change (t2 - t1 = ∆t) 13.5ºC Total mass of solution (assume 1mL = 1 g) 100.1g Heat evolved by reaction 5648.64 J Molarity of NaOH solution 2.0 M Volume of NaOH solution 0.05 L Moles of NaOH 0.1 moles Energy per mole of NaOH 56486.4 J/mole ∆H2 (kJ/mole) NaOH 56.49 kJ/mole Part III Heat of Reaction between Hydrochloric Acid and Solid Sodium Hydroxide Table 3 Original temperature of HCl(aq) (t1) 21.0ºC Final temperature of solution (t2) 34.0ºC Temperature change (t2 - t1 = ∆t) 13.0ºC Mass of HCl (assume 1 mL = 1g) 101.46g Heat evolved by reaction 5513.34 J Mass of NaOH(s) 2.0g Moles of NaOH 0.05001250313 moles Energy per mole of NaOH 110239.23 J/mole ∆H3 (kJ/mole) NaOH 110.24 kJ/mole Questions and Calculations Reaction 1: ∆T=26˚C-21˚C ∆T= 5˚C Cw=4.184J/(g˚C) qw=(100.0g)(4.184J/(g˚C))(26˚C-21˚C) qw=2092J -qrxn=qsolution qrxn=-2092J nNaOH=2.00g×(1 mole)/39.99g=0.05001250313 mols ∆H=qrxn/n=(-2092 J)/(0.05001250313 mols)×1kJ/(1000 J)= ∆H=-41.83 kJ/mole Reaction 2: mass=48.9g+54.2g=103.1g ∆T=34.5˚C-21˚C=13.5˚C Cw=Csolution=4.184J/(g˚C) qsolution=(103.1g)(4.184 J/(g˚C))(34.5˚C-21˚C) qsolution=5823.5 J -qrxn=qsolution
Which of these is the richest source of nitrogen on a mass percentage basis? a) Urea, (NH2)2CO b) Ammonium nitrate, NH4NO3 c) Nitric oxide, NO d) Ammonia, NH3 Cisplatin, an anticancer drug, has the molecular formula Pt(NH3)2Cl2. What is the mass (in gram) of one molecule ? (Atomic weights : Pt = 195, H = 1.0, N = 14, Cl = 35.5) a) 4.98 × 10−21 b) 4.98 × 10−22 c) 6.55 × 10−21 d) 3.85 × 10−22 Aspirin has the formula C9H8O4. How many atoms of oxygen are there in a table spoon weighing 360 mg?
Approximately 3mL of Clear Liquid 2 was carefully poured into the graduated cylinder. The graduated cylinder was placed on the scale again (with the added liquid) and the mass was measured in grams to the nearest .01 gram. The graduated cylinder was then placed on the tabletop and the volume was read to the nearest tenth of a
P.V. is defined as milliequivalent peroxide per 1000g fat. As mentioned, peroxide is produced during lipid oxidation. PV test thus base on the fact that iodine produced by reaction of KI and lipid is proportional to the amount of peroxides generated in oil. ROOH + 2KI I2 + 2KOH + RO- The amount of Iodine produced can be further determined by titration with sodium thiosulphate (NaS2O3) using soluble starch to indicate the endpoint (color change from blue to colorless) I2+NaS2O3 S2O3 + 2NaI After the titrations, we can calculate the P.V.