8. When 38.0 mL of 0.1250 M H2SO4 is added to 100 mL of a solution of PbI2, 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 0.0471g, what was the concentration of iodide ions in the original solution The answer is

Q: Consider the reaction: CaCl2(aq) + 2A9NO3(aq) → Ca(NO3)2(aq) + 2A9CI(s) How many grams of CaCl,…

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Q: When 38.0 mL of 0.1250 M H2SO4 is added to 100. mL of a solution of Pb(NO3)2, a precipitate of PbS04…

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A: moles of AgCl = ( mass / molar mass of AgCl )  moles of AgCl = ( 0.9805g / 143.32gmol-1 )  moles of…

Q: 8. When 38.0 mL of 0.1250 M H2S04 is added to 100 mL of a solution of Pbl2, a precipitate of PbSO4…

A: Given data ;  Volume of H2SO4= 38.0 ml  Concentration of H2SO4= 0.1250 M  Mass of PbSO4= 0.0471 g…

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A: Recall the reaction that will occur here PbNO32 + BaBr2 →PbBr2 + BaNO32 Given that 78.0 ml of 0.524…

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Q: How many grams of PbBr2 will precipitate when excess CoBr2 solution is added to 62.0 mL of 0.530 M…

A: The reaction will be as follows: Pb(NO3)2(aq) + CoBr2(aq)---------------> PbBr2(s) + Co(NO3)2(aq)…

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Q: How many grams of Cu(OH)2 will precipitate when excess KOH solution is added to 41.0 mL of 0.618 M…

A: To solve this problem we will use the given balanced chemical equation.

Q: 4. Consider the mixing of 20.5 mL of 0.500 M NaOH (aq) with 15.2 mL of 0.223 M Ba(NO3)2 (aq). (a)…

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Q: A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further…

A: The reaction of NaCl and Pb(NO3)2 – 2NaCl + PbNO32 →PbCl2 s+ 2NaNO3

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