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.
Calculate the equilibrium constant, Ke. [0.85] 4. If the equilibrium concentration of F2(g) is 1.50 mol/L and H2(g) is 2.5 mol/L, determine the concentration of HF(g) at equilibrium. [1.92 mol/L] F2 (g) + H2 (g) === 2 HF(g) Ke = 0.98 5. If 0.100 mol of hydrogen iodide is placed in a 1.0 L container and allowed to reach equilibrium, find the concentrations of all reactants and products at equilibrium.
Zing Substances. For ionizing substances, such as NaCl , 1mosm is 1mmole times the number of ions formed when each molecule dissociates. One mmole of NaCl is 58 mg, but when it dissociates, it yields 1 mmole of Na+ (23mg) and 1mmole of Cl ( 35mg). Therefore , 58 mg of NaCl is 2 mOsm of NaCl is put into a beaker and distilled water added to make 1 liter, the osmolarity is 2mOsm/l. A) How many mosm solute will 1 gram of NaCl yield?
Thus, the molarity of the HCl solution can be calculated by dividing the number of moles of HCl by the volume of HCl (in liters) used to neutralize the Na2CO3 . Now that it is a neutralized solution, we are able to use it for the titration of NaHCO3. NaHCO3(aq.) + HCl(aq.) ==> NaCl(aq.)
White precipitate shows the presence of chloride (Cl-). Chloride anion equation: HCl(aq) + AgNO3 (aq) → HNO3 (aq) + AgCl(s). The nitrate anion test involves cooling a mixture containing 1 mL of test solution and 3mL 18M H2SO4. 2mL is poured down the inner test tube side and the presence of a brown ring shows nitrate (NO3-) to be present. The carbonate anion test mixes 1 mL of test solution and drops of 6M HCl.
A. 0.420 M B. 0.567 M C. 0.042 M D. 0.325 M _____ 3. What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose, C6H12O6, per 100.0 g of H2O ? (Kf for water = 1.86 °C /m) A.
Chemistry: Identifying Chemical Reactions Name_____________________________________________________ Date___________________________ Your text book identifies four different types of chemical reactions (section 16-4). Other sources identify as many as six types of chemical reactions. For now we will stick with the scheme used by your book. The four types of chemical reactions shown in your book are: a) Synthesis b) Decomposition c) Single displacement or Single replacement d) Double displacement or Single replacement e) Combustion Write balanced equations for each of the word equations below, then indicate which kind of reaction it is 1. Sodium bromide + Calcium hydroxide
Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3. The general formula of barium chloride hydrate is BaClg-nHZO, where n is the number of water molecules. Calculate the theoretical percent water for each value of n—divide the sum of the atomic masses due to the water molecules by the sum of all the atomic masses in the hydrate, and multiply the result by 100. Complete the table. | BaCl2 | BaCl2•H2O | BaCl2•2H2O | BaCl•3H2O | Sum of atomic masses (BaCl2) | 208.23 | 208.23 | 208.23 | 208.23 | Sum of atomic masses (nH2O) | 0 | 18.02 | 36.04 | 54.06 | Sum of atomic masses (hydrate) | 208.23 | 226.25 | 244.27 | 262.29 | Percent water in hydrate (theoretical) | 0% | 7.96% | 14.75% | 20.61% | In this lab we used a Balance, centigram
1. Equations for each reaction in the preparation. (a) Reaction of copper (II) oxide with nitric acid. CuO (s) + 2HNO3 (aq) → Cu(NO3)2 (aq) + H2O (l) (b) Neutralization of the excess nitric acid with sodium carbonate. Na2CO3 (s) + 2HNO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l) (c) Reaction of aqueous copper(II) nitrate with sodium carbonate.