1)
Mol of H+ from HCl = M(HCl)*V(HCl) in L
= 0.340 M * 0.110 L
= 0.0374 mol
Mol of H+ from HNO3 = M(HNO3)*V(HNO3) in L
= 0.150 M * 0.330 L
= 0.0495 mol
Total mol of H+ = 0.0374 mol + 0.0495 mol
= 0.0869 mol
Answer: 0.0869 mol
Only 1 question at a time please
1) A 110 mL solution of 0.340 M HCl(aq) is mixed with a solution of 330...
H+ are present in the final SulULUI 2) When 38.0 mL of 0.1250 M H2SO4 is added to 100 mL of a solution of Pbl2, a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution, dried, and weighed. If the recovered PbSO4 is found to have a mass of 0.0471 g, what was the concentration of iodide ions in the original solution? Jad ta neutralize 50.0 mL of an H2SO4 solution.
2) When 38.0 mL of 0.1250 MH,SO, is added to 100 mL of a solution of Pbly, a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution, dried, and weighed. If the recovered PbSO4 is found to have a mass of 0.0471 g, what was the concentration of iodide ions in the original solution?
2) When 52.0 mL of 0.1250M H2SO4 is added to 100 mL of a solution of Pbl2, a precipitate of PbSO4 forms. The PBSO4 is then filtered from the solution, dried, and weighed. If the recovered PbSO4 is found to have a mass of 0.0471 g, what was the concentration of iodide ions in the original solution?
Gravimetric Analysis.. How did they get this answer When of a 0.3000 M AgNO_2 solution is added to 50.0 mL of a solution of MgCl_2, an AgCI precipitate forms immediately. The precipitate is then filtered from the solution, dried, and weighed. If the recovered AgCI is found to have a mass of 0.1183_g, what was the concentration of magnesium ions in the original MgCl_2 solution?
When 20.0 ml of a 0250 M (NHS solution is added to 1500 ml of a solution of CuNOJ, CuS preciptate forms The precipitate is then filtered from the solution, dried, and weighed If the recovered CuS is found to have a mass of 0.3491 g what was the concentration of copper lons in the original CuNO) solution? 333-102 M O 365 10 M O 243 102 M O 122 102 M O 487 102 M
(III) When 50.0 mL of a 0.3000 M AgNO, solution is added to 50.0 mL of a 0.2000 M solution of MgCl, an AgCl precipitate forms immediately. The precipitate is then filtered from the solution, dried, and weighed. If the recovered AgCl is found to have a mass of 1.7825 g, what is the percentage yield of the product, 1) Write the balanced equation for the reaction. 2) Calculate number of moles of each reactant and determine which one is...
A 55.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead (II) acetate solution. The solid lead (II) sulfate is collected, dried, and found to have a mass of 1.01 g. The Balanced Reaction is: Pb(CH3COO2)(aq) + K2SO4 --> PbSO4(S) + 2K (aq) + 2CH3COO- (aq) b.Write the net ionic reaction. c.Which ions are spectators? d.Determine the limiting reagent. e. Determine the theoretical yield. f.Determine the percent yield. g. Determine...
65.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 38.5 mL of a 0.122 M lead(II) acetate solution and the following precipitation reaction occurs: K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 0.993 g . a. Define the theoretical yield b. Define the percent yield.
A 77.0-mL sample of a 0.203 M potassium sulfate solution is mixed with 55.0 mL of a 0.226 M lead(II) nitrate solution and this reaction occurs: K2SO4(aq) + Pb(NO3)2 (aq) ⟶ 2 KNO3(aq) + PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 3.71 g. Determine the percent yield. 1 mole PbSO4 = 303.26 g Group of answer choices 98.4% 80.0% 120.% 75.7%
A 77.0-ml sample of a 0.203 M potassium sulfate solution is mixed with 52.0 mL of a 0.214 Mlead(11) nitrate solution and this reaction occurs: K2SO4(aq) + Pb(NO3)2 (aq) 2 KNO3(aq) + PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 2.43 g. Determine the percent yield. 1 mole PbSO4 = 303.26g 80.1% O 120% o 58.9% o 72.0%