A 0.450 g sample of impure CaCO, (s) is dissolved in 50.0 mL of 0.150 M...
A 0.450 g sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq) . The equation for the reaction is CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) The excess HCl(aq) is titrated by 9.05 mL of 0.125 M NaOH(aq) . Calculate the mass percentage of CaCO3(s) in the sample.
< Question 1 of 3 > A 0.450 g sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO3(s) + 2 HCl(aq) → CaCl, (aq) + H2O(1) + CO2(g) The excess HCl(aq) is titrated by 8.45 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO,(s) in the sample. mass percentage:
A 0.450 gram sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO3(s) + 2HCl(aq) arrow CaCl2(aq) +H2O(l) +CO2(g) The excess HCL (aq) is titrated by 6.45 mL of 0.125 M NaOH(aq). Calculate the mass percentaageof CaCO3(s) in the sample. Please show steps.
The zinc content of a 1.23 g ore sample was determined by dissolving the ore in HCI, which reacts with the zinc, and then neutralizing excess HCl with NaOH. The reaction of HCI with Zn is shown below. Zn(s)-2HCl(aq) → ZnCl2(aq)-H,(g) The ore was dissolved in 150 mL of 0.600 M HCI, and the resulting solution was diluted to a total volume of 300 mL. A 20.0 mL aliquot of the final solution required 9.19 mL of 0.529 M NaOH...
The zinc content of a 1.45 g ore sample was determined by dissolving the ore in HCl, which reacts with the zinc, and then neutralizing excess HCl with NaOH. The reaction of HCl with Zn is shown below:Zn+2HCL--ZnCl2+H2The ore was dissolved in 150 mL of 0.600 M HCl, and the resulting solution was diluted to a total volume of 300 mL. A 20.0 mL aliquot of the final solution required 8.48 mL of 0.508 M NaOH for the HCl present...
Here is the question: picture is below for clarity. The zinc content of a 1.301.30 g ore sample was determined by dissolving the ore in HClHCl, which reacts with the zinc. The excess HClHCl is then neutralized with with NaOHNaOH. The reaction of HClHCl with ZnZn is shown.Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)The ore was dissolved in 150 mL150 mL of 0.600 M HCl0.600 M HCl, and the resulting solution was diluted to a total volume of 300 mL300 mL. A 20.0 mL20.0 mL aliquot of the final solution required 9.429.42 mL of 0.5100.510 M NaOHNaOH to neutralize the excess HClHCl. What is the mass percentage (%w/w) of ZnZn in the ore sample?
5. A 2.600 g sample of limestone was dissolved in excess 0.200 M HCl(aą): CaCO(s) 2 HCa)Ca (a2 CT(a)CO (g)HO) Excess 0.10 M (NH C20(a) was added to the resulting solution to precipitate the calcium ions as calcium oxalate, CaC204(s). Ca (aq) (NH).CO(a)CaC,04(s)2 NH (aq) The precipitate was filtered and dried three times to a constant mass of I 036 g. Determine the mass% Ca in the limestone sample.
A 0.6739 g sample of a pure carbonate, X,CO,(s), was dissolved in 50.0 mL of 0.1850 M HCl(aq). The excess HCl(aq) was back titrated with 24.70 mL of 0.0980 M NaOH(aq). How many moles of HCl react with the carbonate? moles of HCI = mol What is the identity of the cation, X? cation: A standardized solution that is 0.0100 M in Na+ is necessary for a Hame photometric determination of the element. How many grams of primary-standard-grade sodium carbonate...
1. A 0.7336-g sample of an alloy containing copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH4PO4, and subsequently isolated as Zn2P2O7, yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.1931 g. Calculate the %w/w Zn and the %w/w Cu in the sample.
A sample of copper ore with a mass of 0.4225 g was dissolved in acid. A solution of potassium iodide was added, which caused the reaction: 2Cu2+(aq) + 5I– (aq) → I3–(aq ) + 2CuI(s) The I3– that formed reacted quantitatively with exactly 29.96 mL of 0.02100 M Na2S2O3 according to the following equation: I3– (aq) + 2S2O32– (aq) → 3I– (aq ) + S4O62–(aq ) What was the percentage by mass of copper in the ore? If the ore...