Chemistry question? Solutions?
A solution is made by dissolving 19.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δtf of the solution? Show all he steps taken to solve this problem.
Ahhh! solutions??? chemistry question?
Strong acid. Hydronium ions are H3O+ ions, H+ ions are made from strong acids (they dissociate fully).
Chemistry Question: Solutions?
This is what the answer is, according to the teacher's Key. I don't really understand what is going on: 4. Describe the preparation of 0.30 L of a 8.2% ammonium sulfate solution. Percent by mass = grams of solute x 100 grams of solution 8.2 % = ____X g_______ x 100 X g + 3.0 x 102 g 0.082 = ____X g_______ X g + 3.0 x 102 g 0.082X + 0.082(3.0 x 102) = X X = 27 g Percent by mass = grams of solute________ x 100 grams of solute + grams of solvent 0.082(3.0 x 102) = X – 0.082X Place 27 g of ammonium sulfate into a container and add 300 mL of water.
AP Chem question, solutions?
V (Na2CO3) = 0,070 L V (NaHCO3) = 0,030 L V (total) = 0,070 L + 0,030 L = 0,100 L [Na2CO3] = 3,0 mol/L [NaHCO3] = 1,0 mol/L n (Na2CO3) = 3,0 mol/L * 0,070 L = 0,210 mol n (NaHCO3) = 1,0 mol/L * 0,0,30 L = 0,030 mol The ratio between Na2CO3 and NaHCO3 is 7:1: 7Na2CO3 + NaHCO3 -> Na+ (+ CO3(2-) + HCO3-) Balanced: 7Na2CO3 + NaHCO3 -> 15Na+ n (Na+) = 15 * (n (NaHCO3)) = 15 * 0,030 mol = 0,450 mol [Na+] = n (Na+) / V (total) = 0,450 mol / 0,100 L = 4,50 M Of course the poster above me is absolutely correct, and this is just another way of doing it.
Chemistry question on volume of solutions?
A solution is 74.5% by mass of substance A. The density of the solution is 1.37 g/ml. A reaction requires 5.23 g of A. What volume of solution is required? Please show work.
Chemistry question (properties of solutions)?
Concentrated sulfuric acid (18.4 molar H2SO4) has a density of 1.84 grams per milliliter. After dilution with water to 5.20 molar, the solution has a density of 1.38 grams per milliliter and can be used as an electrolyte in lead storage batteries for automobiles. a) Calculate the volume of concentrated acid required to prepare 1.00 liter of 5.20 molar H2SO4. b) Calculate the volume of 5.20 molar H2SO4 that can be completely neutralized with 10.5 grams of Sodium Bicarbonate, NaHCO3. c) What is the molality of the 5.20 molar H2SO4. Show work please so I can learn. THANK YOU VERY MUCH!
What is the solution to the following chemistry question?
Correct answer is option 3. O3. As Srinath mentioned the process is called ozonolysis. For mechanism see this video Halogenation and ozonolysis of alkynes
AP Chemistry question on solutions! URGENT HELP!?
A) 49 g of H2SO4 = 0.5 moles. In order to prepare a 1.0 M solution, you would need to make 0.5 liters of solution (0.5 moles/0.5 liters = 1.0 moles/liter). Dissolve the H2SO4 in water and dilute to 500 mL. B) A 1.0 molal solution is equivalent to moles solute/kg solvent. In this example, you have 0.5 moles of H2SO4 (your solute) so you would need to dissolve that in 0.5 kg water (your solvent). In this case, your total volume is going to be slightly more than 500 mL, since the volume of H2SO4 is going to add to the total. C) Which gives you a greater % H2SO4? 49 g H2SO4 with water added to make a total of 500 mL, or 49 g H2SO4 with 500 mL water added? D) Mixing a strong acid with water generates significant quantities of heat (enough to actually boil the water, if there's not enough of it). Acid must be added to sufficient water. Never add water to concentrated acid. E) The energy change involved with breaking apart a crystal lattice is always endothermic. In an endothermic dissolving process, the crystal lattice energy is going to be more endothermic than the energies of hydration.