There will have some error. 2) A volatile liquid was allowed to evaporate in a 43.298 g flask that has a total volume of 252 ml. the temperature of the water bath was 100˚C at the atmospheric pressure of 776 torr. The mass of the flask and condensed vapor was 44.173 g. calculate the molar mass of the liquid. T = 273 + 100 = 373 V = 252 mL = 1 L / 1000 mL = 0.252 L P = 776 Torr R= 0.0821 mass of 44.173 - 43.298 g = 0.875g moles of gas = PV / RT = 776 x .252 / 62.363 x (273+100) =0.00841 moles molar mass = 0.875g / 0.00841 moles = 104.1 g/
A) is neutralized by water B) is surrounded by water molecules C) reacts and forms a covalent bond to water D) aggregates with other molecules or ions to form a micelle in water Answer: B Page Ref: Section 3 11 9) Which would you expect to be most soluble in water? A) I B) II C) III D) IV Answer: A Page Ref: Section 3 10) Solutes diffuse more slowly in cytoplasm than in water because of A) the higher viscosity of water. B) the higher heat of vaporization of water. C) the presence of many crowded molecules in the cytoplasm. D) the absence of charged molecules inside cells.
The following mistakes were made when carrying out the experiment. What effect does each have on the calculated molar mass? Be specific. For example, too large because… Only part of the pipet was immersed in the boiling water, so the temperature in part of the pipet was less than that of the water bath. If the temperature was less than the water bath in some places because only part of the pipet was immersed in the boiling water, the molar mass calculated would become lower.
93 g/mol? Not we get to use it! Yay! 93 g/mol / 31.06 g/mol = 3 (this is the multiplier) Multiply that whole number through the subscripts of the empirical formula. 3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off.
If there are lower levels of potassium, then the resting membrane potential will be higher than normal and an action potential will be more likely with less stimulation. Less voltage will need to be used. 2. Tetrodotoxin, a toxin found in puffer fish, acts by inhibiting voltage-gated sodium channels. Eating improperly prepared puffer fish sushi can be fatal because of interference with action potential generation.
This solution was placed in a burette and 18.4 cm3 was required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Deduce the molecular formula of the acid and hence the value of n. 5. Sodium carbonate exists in hydrated form, Na2CO3.xH2O, in the solid state. 3.5 g of a sodium carbonate sample was dissolved in water and the volume made up to 250 cm3. 25.0 cm3 of this solution was titrated against 0.1 moldm-3 HCl and 24.5 cm3 of the acid were required.
We used 1g of Na2CO3 -Sodium Carbonate. This would allow for the full precipitate to form. 4) Procedure: Refer to CHE 111 lab manual, experiment Stoichiometry of a Precipitation Reaction on page #104 5) Data / results. We used 1g of Na2CO3 -Sodium Carbonate in combination with 1g CaCl2·2 H2O-Calcium Chloride, Dihydrate to form the precipitate reaction. Na2CO3 aq + CaCl2*2H2O aq = CaCO3 s + 2 NaCl aq + 2 H2O aq Molar Mass: Ca = 40.08 Cl2 = (35.45) x 2 = 70.9 2H2O = (18.02) x 2 = 36.04 40.08 + 70.90 + 36.04 = 147.02 g/mol CaCl2*2H2O Theoretical yield: 1g CaCl•2H2O x 1 mol CaCl•2H2O x 1 mol CaCO3 x 100.09g CaCO3 = 0.68 g CaCO3 147.02g CaCl•2H2O 1 mol CaCl•2H2O 1 mol CaCO3 Mass of weighing dish = 0.6g Mass of filter paper =
The percent yield was calculated to be 64.5%. Overall, this experiment was pretty efficient. One error that could have affected the actual yield is that when pouring the solution into the filter paper, some of the solution could have spilled out or when finished pouring the solution, some of the precipitate was still stuck at the bottom of the flask. Secondly the filter paper might not have been folded correctly. If that were to happen, some of the precipitate would have escaped through the paper and into the second
The freezing point depression has some limitations and factors which affect to what extent the freezing point of the solvent will be decreased. One major limitation is the choice of the solute to be dissolved in the solvent because the solute must be able to dissolve sufficiently to lower the freezing point of the solvent. The solute must be soluble in the solvent in order to form a homogenous mixture. A few experiments with the unknown and different solutes will tell you what kind of properties the substance has and what substances will dissolve in it. Also, there is a limit of how much solute you can put in the solvent and you should not exceed that certain percentage.
The specific heat constant for water, 4.184 J/g C, is used for this equation. The specific heat can be found by using The Law of Dulong and Petit: Eq. 3 Cs(aluminum) = slope x 1/atomic weight This equation is used to find specific heat from the graph that will be drawn based on the results of the metal specific heats. II. Materials and Procedure See General Chemistry 101/102 Laboratory Manual (pg.