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.
[7.7(10-12 mol/L] 2. How many milligrams of PbI2 can you dissolve in 300 mL of water at 25(C? Ksp of PbI2 is 1.4(10-8 [210 mg] 3. Calculate the solubility product constant, Ksp, of SrCl2 if 8.0 mg dissolves in 200 mL to form a saturated solution at 25(C. [6.4(10-11] 4. Seawater is saturated with AgCl.
(57.10) 7 in. 3. What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has total equivalent length of 50 ft? (57.10) 0.2 4. What size metal duct should be used to deliver 270 CFM with a pressure drop of 0.15 in wc if the total equivalent length is 80 ft?
What volume of 2.0 M hydrochloric acid is needed to completely react with the amount of calcium carbonate in Part 2a above? c. Based on Parts 2a and 2b above, how many moles of water would be produced? 3. Ammonium chloride and calcium hydroxide react according to the following balanced equation: 2 NH4Cl(aq) + Ca(OH)2(aq) ⋄ CaCl2(aq) + 2 NH3(g) + 2 H2O(l) a. What mass of ammonium chloride is needed to make 3.0 liters of a 1.5 M ammonium chloride solution?
c. Prepare the solution by dissolving 38.90 grams of ZnI2 with 500 mL of water. d. 0.0125/0.25 = 0.05 L = 50 mL. This produces 0.0125 moles of ZnI2 5. Exercise 5: a. (0.125)(0.1) = 0.0125 moles of solute b. Pour 50 mL of the stock solution to get the number of moles needed.
3. How many moles of hydrogen gas can be produced when 10.0 g of Zn react with 50.0mL of 2.0M HCl? 4. If the pressure is 1.2 atm and the temperature is 20.0ºC, what volume of hydrogen is produced in prelab question #3? Procedure: 1.
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.
Whenever you record the volume, record to the nearest 0.01mL. Do this six times. Data: Trial # Volume of water after addition of metal m(L). Mass of the container and metal (g). 1 5.24 mL 10.061 g 2 5.49 mL 9.293 g 3 5.99 mL 8.055 g 4 6.51 mL 6.647 g 5 7.02 mL 5.189 g 6 7.51 mL 3.932 g Calculation: Error: Our percent error was -0.37%.
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 the reaction is not spontaneous under standard conditions at 298K, at what temperature (if any) would the reaction become spontaneous? a) 2 PbS (s) + 3 O2 (g) → 2 PbO (s) + 2SO2 (g) ; ΔH° = -844 kJ; ΔS°= -165 J/K b) 2 POCl3 (g) → 2 PCl3 (g) + O2 (g) ; ΔH° = 572 kJ; ΔS°= 179 J/K 5. Consider the reaction H2 (g) + F2 (g) → 2 HF (g). a) Using data in your Appendix B, calculate ΔG° at 25°C b) Calculate ΔG at 298K if the reaction mixture consists of 8.0 bar of H2, 4.5 bar of F2 and 0.36 bar of HF. 6.