(2 points) Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq) 2. Determine the partial pressure of the hydrogen gas collected in the gas collection tube. (3 points) partial pressure H2 = total pressure - vapor pressure of water = 746mmHg - 19.8mmHg = 726mmHg 3. Calculate the moles of hydrogen gas collected. (4 points) n = 125 4.
Calculate the value of Ke for this system. 2 H2S (g) === 2 H2 (g) + S2 (g) [1.1(10-4] 7. At a given temperature, the following system has an equilibrium constant, Ke, of 0.27. C(g) + B(g) === 2 E(g) The system was established by placing 8.00 moles of C and 5.0 moles of B in a 4.0 L vessel. Calculate the concentration of all substances at equilibrium.
20. mol H2 reacts with 8.0 mol O2 to produce H2O. Determine the number of grams reactant in excess and number of grams H2O produced. Identify the limiting reactant. 8.1 g H2 , 2.9 x 102 g H2O 17. How many litres of O2 gas are required to produce 100. g Al2O3?
Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium. The reaction is as follows: Mg + 2HCl -> H2 + Mg2+ (aq) + 2Cl- (aq) There is a one to one relationship between the number of moles of hydrogen gas evolved and the
Name___________________________________ periodic table review 3 12/5/2006 Page 1 Key Terms 1. What is the name of the family of elements in Group IA/1? a. alkali metals b. alkaline earth metals c. halogens d. noble gases e. none of the above 2. What is the name of the family of elements in Group IIA/2? a. alkali metals b. alkaline earth metals c. halogens d. noble gases e. none of the above 3.
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. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.
The one peak at 3.482 ppm indicates the –OCH3 group, because the hydrogen atoms (HA) are adjacent to the highly electronegative oxygen atom. The –CH3 hydrogen atoms observe no splitting because they have no neighboring protons (N=0) therefore it is a singlet. The three peaks at around 0.90 ppm represent the –CH3 group on the opposite end of the molecule, because the three hydrogen atoms (HB) are furthest from the electronegative oxygen atoms and because the protons have two neighboring protons (N=2) indicating a triplet. The three peaks at around 2.40 ppm represent the –CH3 group adjacent to the C=O bond, because the hydrogen atoms (HC) have two neighboring protons (N=2) indicating a triplet and it is next to the electronegative double bond. The NMR spectrum does contain impurities including methanol (4.80 ppm), methyl oleate (5.40 ppm), CHCL3 (7.20 ppm), acetone (2.00 ppm), and water (1.60 ppm).
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/
The following data were obtained when a sample of barium chloride hydrate was analyzed as described in the Procedure section. Calculate (a) the mass of the hydrate, (b) the mass of water lost during heating, and (c) the percent water in the hydrate. Mass of empty test tube 18.42 g Mass of test tube and hydrate (before heating) 20.75 g Mass of test tube and anhydrous salt (after heating) 20.41 g. Mass of the Hydrate is 2.33g. Loss (H2O) is 0.34g. Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3.
12 C. 5 D. 30 _____ 16. A voltaic cell has E0cell = +1.00 V. The cell reaction A. is not spontaneous B. has K = 1 C. has (G0 = 0 D. has a negative (G0 _____ 17. Which energy conversion takes place in a galvanic cell? A. electrical to chemical B. chemical to