Determine the mass (in g) of Co(OH)2 that is produced when 974 mL of a 1.20x102...
Determine the mass (in g) of Ba3(PO4)2 that is produced when 125 mL of a 4.55×10-2 M Ba(NO3)2 solution completely reacts with 519 mL of a 6.92×10-2 M Na3PO4 solution according to the following balanced chemical equation. 3Ba(NO3)2(aq) + 2Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq)
1.What mass of HKC8H4O4 reacts with 25.00 mL of NaOH a solution containing 0.800 g NaOH / L.? 2.A 0.3396 g sample of pure 96.4% Na2SO4 reacts quantitatively with 37.70 mL of a barium chloride solution. The reaction is: Ba + 2 (ac) + SO4-2 (ac) → BaSO4 (s) Calculate the molar concentration of barium ion in the solution. 3. What mass of Ag2CO3 is obtained from the reaction of 125 mL of 0.500 M AgNO3 with about 125 mL...
In the laboratory, a student dilutes 26.4 mL of a 10.5 M nitric acid solution to a total volume of 250.0 mL. What is the concentration of the diluted solution? Concentration = How many milliliters of 11.6 M hydrochloric acid solution should be used to prepare 5.50 L of 0.100 M HCl? Interpret the following equation for a chemical reaction using the coefficients given: CO(g) + Cl2(g) →→ COC12(g) On the particulate level: of CO(g) reacts with — of Cl2(g)...
8. When glucose reacts with oxygen in living systems, carbon dioxide and water are produced, and a great deal of energy is liberated. C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) What weight of carbon dioxide can be produced from the reaction of 10.0 grams of glucose with 10.0 grams of oxygen? (a) 2.29 g (b) 2.44 g (c) 13.8 g (d) 14.7 g (e) none of these 9.Based on the balanced chemical equation shown below, determine the...
Consider the following reaction. MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq) A 166.0 mL solution of 0.381 M MgCl2 reacts with a 47.33 mL solution of 0.568 M NaOH to produce Mg(OH)2 and NaCl. Identify the limiting reactant. NaOH Mg(OH)2 MgCl2 NaCl Caclulate the mass of Mg(OH)2 that can be produced. The actual mass of Mg(OH)2 isolated was 0.559 g. Calculate the percent yield of Mg(OH)2.
49.50 mL of a 2.74 M NaOH solution reacts completely with 40.44 mL of HCI solution according to the balanced chemical reaction shown below: HCI (aq) + NaOH(aq) - NaCl(aq) + H2O(1) How many moles of HCl reacted?! Please round your answer to the thousandths place. Your Answer:
Consider a chemical reaction CO(g) + 2H, (g) = CH OH(g). When 2 moles of Co were reacting with 4 moles of H, at temperature 500K and pressure p = 100 bar, and chemical equilibrium was reached, n = 0.366 moles of CO was consumed. a) (10 pts) Based on the given data determine the value of the chemical equilibrium constant K(T) for T = 500K. b) (10 pts) Consider now the same reaction, when 3 moles of CO are...
Prelab Activity: Titrations Continued – Titration of Household Products A 2.40 g sample of vinegar was added to an Erlenmeyer flask along with 100 mL of deionized water and 3 drops of phenolphthalein indicator. It took 22.15 mL of 0.0981 M NaOH (aq) to reach the faint pink endpoint. The following balanced chemical equation represents the chemistry of the titration: NaOH (aq) + CH3COOH (aq) à CH3COO–Na+ (aq) + H2O (l) Calculate the mass % of acetic acid (CH3COOH) in...
N2(g)3H2(g)2NH3(g) Answer Consider the following balanced chemical equation 4KO2(s)2H20(I)302(8) +4KOH(s) Determine the mass (in g) of (a) KOH formed if 10.0 g of KO2 reacts with 10.0 g of H2O. Identify the limiting reactant. Determine the mass (in g) of KOH formed when 20.0 g of (b) KO2 reacts with 10.0 g of H20. Identify the limiting reactant. Determine the mass (in g) of (c) O2 formed when 25.0 g of KO2 reacts with 5.00 g of H20. Identify the...
2.4.0 mL of 3.00 M HCl reacts completely with the CaCO3 (active ingredient) in an antacid tablet. The remaining solution (excess HCl) is then titrated with 0.0981 M NaOH to the bromothymol blue endpoint. Write the balanced chemical equation for the reaction of CaCO3 (s) with HCl (aq). If the volume of NaOH delivered at the endpoint is 22.20 mL, determine the amount (in milligrams) of CaCO3 that was contained in the antacid tablet.