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.
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.
1 / [CO2] C. [CaO][CO2] / [CaCO3] D. [CaCO3] / [CaO][CO2] _____ 13. The value of Kp for the reaction 2 NO2 (g) [pic] N2O4 (g) is 1.52 at 319 K. What is the value of Kp at this temperature for the reaction N2O4 (g) [pic] 2 NO2 (g) ? A. -1.52 B. 1.23 C. 5.74 X 10-4 D. 0.658 _____ 14.
0.00512 * 500 = 2.56 c. 0.00806 * 319.9= 2.56 d. 0.00806 * 319.9 = 2.56 500 mL =0.5 L e. 0.0161 * 500 / 2 =2.56 4. Exercise 4: a. 0.250M; 250 mL 0.25 L (0.25) (0.25) = 0.0625 moles ZnI2 b. Prepare the solution by dissolving 19.95 grams of Zinc Iodide with 250 mL of water. c. Prepare the solution by dissolving 38.90 grams of ZnI2 with 500 mL of water.
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
5. Potassium hydroxide solution: Use 0.01 M potassium hydroxide, 0.56 g/L. 6. Ammonium hydroxide solution: Add water to 10 mL of concentrated ammonium hydroxide solution to make 82 mL of a stock solution. Use 10 mL of the stock and dilute to 1 L with distilled water.
0.00512g ZnI2/mL of solution 0.00512g/319.18 g/mol=1.6*10-5 mol 1.6*10-5 mol/(1*10-3L)=0.016M c. 0.00806 moles of ZnI2/500 mL of solution 0.00806mol/(500*10-3)L=0.016M d. 0.0161 moles of ZnI2/L of solution 0.0161mol/1L=0.016M Exercise 4: a. The moles of ZnI2: 0.25M*(250*10-3)L=0.0625mol b. 0.25M*(250*10-3)L=0.0625mol The mass of ZnI2: 0.0625mol*319.18 g/mol=19.95g c. 0.25M*(500*10-3)L=0.125mol 0.125mol*319.18 g/mol=39.9g ZnI2 d. 0.0125mol/0.25M=0.05L Exercise 5: a. 0.125M*(100*10-3)L=0.0125mol b. 0.0625mol/0.125M=0.5L=500mL Calculation for preparing the EDTA solution Exercise 6 a.
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.
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/