u.S.00V 10. What mass of AgCl is precipitated from the reaction of 25.0 mL of 0.100M AgNO3 and 100. ml 0.0500M KCI? AgCl(s)+ KNO3(aq) AgNOs(aq) + KCl(aq) a. 0.0050 g b. 0.355 g d. 0.425 g c. 0.717 g
Q: A 0.3000 g sample of impure Na2CO3 requires 36.31 mL of a 0.1075 M HClO4 solution to reach the end…
A: Molarity of HClO4 = 0.1075 M Volume of HClO4 = 36.31 mL = 0.03631 L Mass of sample = 0.3000 g We…
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A: The given chemical equation: AgNO3 + NaCl → AgCl + NaNO3 1 mol 1 mol 1 mol 1 mol…
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Q: Given the reaction: 3K2CO3 +2H3PO4(aq) → 2K3PO4(aq) +3CO2(g) +3H2O(l) If it required 97.65 ml…
A: Given: The balanced reaction is 3K2CO3 +2H3PO4(aq) → 2K3PO4(aq) +3CO2 (g) +3H2O(l) Volume of H2PO4 ,…
Q: A solution of 25 mL Na2CO3 is neutralized with 22.4 mL of 0.1065 M HCl. In the another analysis, 50…
A: Given that, Volume of 0.1065 M HCl = 22.4 mL = 0.0224 L Moles of HCl in 0.0224 L = 0.1065 M×0.0224…
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A: Given data,Molarity of Mn2+=3.25×10-2MVolume of Mn2+ solution=0.654LMolarity of Na2CO3=0.742MVolume…
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A: Given: Volume of nitric acid=35.0 mL Molarity of nitric acid=0.255 M Volume of Mg(NO3)2=0.328 M To…
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Q: Calculate the following when 0.250 L of 0.213 M AgNO3 solution is mixed with 0.350 L of 0.300 M…
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Q: How many grams of Ag2CO3 will precipitate when excess (NH4)2CO3 solution is added to 44.0 mL of…
A: Given, the chemical reaction as follows, 2AgNO3(aq) + (NH4)2CO3(aq) => Ag2CO3(s) + 2NH4NO3(aq)…
Q: A solution of 118mL of 0.180 M KOH is mixed with a solution of 260mL of 0.190M NiSO4. = 2KOH(aq) +…
A: Volume of KOH = 118mL Molarity of KOH = 0.180 M KOH Number of moles of KOH = 118 × 0.180/1000 =…
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Q: A 5.0 g impure sample of sodium carbonate was titrated with an acid. The sample utilized 31.55 ml of…
A: Given as MM of Na2CO3 = 106 g/molMM of HCl = 36.46 g/mol Mass of Na2CO3=5.0 g Volume of HCl…
Q: B. Use the equation to calculate the following: K,CO3(aq) + 2 HC,H¿O,(aq) → 2 KC,H¿O¿(aq) + H,O(!) +…
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A: First of all we will calculate the moles of AgNO3 Moles of AgNO3 = 2.50 L X 0.200 M = 0.5 mole Here…
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A: Mass of AgNO3 = 0.4522 g Volume = 200 mL Mass of K2CrO4 = 0.1668 g Volume = 100 mL Precipitates…
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A: The balanced equation for the titration of oxalic acid with potassium permanganate--2 KMnO4+ 5…
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- 95. Many metal ions form insoluble sulfide compounds when a solution of the metal ion is treated with hydrogen sulfide gas. For example, nickel(II) precipitates nearly quantitatively as NiS when H2S gas is bubbled through a nickel ion solution. How many milliliters of gaseous H2S at STP are needed to precipitate all (he nickel ion present in 10. mL of 0.050 M NiCl2 solution?Nickel(II) formate [Ni(HCO2)2] is widely used as a catalyst precursor and to make metallic nickel. It can be prepared in the general chemistry laboratory by treating nickel(II) acetate with formic acid (HCO2H). Ni(CH3CO2)2(aq) + 2 HCO2H(aq) Ni(HCO2)2(aq) + 2 CH3CO2H(aq) Green crystalline Ni(HCO2)2 is precipitated after adding ethanol to the solution. (a) What is the theoretical yield of nickel(II) formate from 0.500 g of nickel(II) acetate and excess formic acid? (b) Is nickel(II) formate paramagnetic or diamagnetic? If it is paramagnetic, how many unpaired electrons would you expect? (c) If nickel(II) formate is heated to 300 C in the absence of air for 30 minutes, the salt decomposes to form pure nickel powder. What mass of nickel powder should be produced by heating 253 mg of nickel(II) formate? Are nickel atoms paramagnetic?A vinegar solution was prepared by diluting 25.00 mL of vinegar to 250.0 mL. From this solution, 50.00 mL portion was taken, and this required 30.00 mL of 0.1000M NaOH solution to reach neutralization end point. What is the percentage (w/v) of HOAc in the sample? Mol wt HOAc = 60.0 g/mol
- The mass of K3PO4 needed to prepare 250.0mL of an aqueous solution in which PO4-3 concentration is 0.0550M. The answer is ……………………… How many grams of silver sample is equal to 0.0417 mole of silver The answer is ……………………… 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 ……………………………… Express 96.342 m using 2 significant figures The answer is ………………………………. The oxidation number of sulfur in (Na2S2O5) is? The answer is ………………………………Calculate the weight of pure sodium carbonate that is necessary to prepare 2.806 L of 0.223 N Na2CO3 (105.99 g/mol) from the primary-standard solid. Assume the solution is to be used for titrations in which the reaction is: CO32- + 2H+ ----> H2O + CO2 Express you answers in 3 decimal placesThe solutbility constant for Ce(IO3)3 is 3.2x10^-10. What is the Ce3+ concentration in a solution prepared by mixing 50 mL of 0.0450 M Ce3+ with 50 mL of: A.) 0.0450 M IO3- ? B.) 0.0500 M IO3- ?
- A sample of pure sodium oxalate weighing 0.1050 g is ignited [Na2C2O4 (s) --> Na2CO3 (s) + CO (g)] and the resulting product requires 15.00 mL of a solution of H2SO4 for complete neutralization. What is the normality of the acid? MM of Na2CO3 = 106.0 g/molMM of Na2C2O4 = 134.0 g/molMM of CO = 28.01 g/molMM of H2SO4 = 98.08 g/molIf 100.0 mL of 0.167 M Na2SO4 is added to 100.0 mL of 0.897 M Pb(NO3)2, How many grams of PbSO4 can be produced? ??2??4(??) + ??(??3)2(??) → 2????3(??) + ????4(?)A solid sample containing some Fe2+ion had a total mass of .9791 g. It required 18.2 ml of 0.02304 M KMnO4 to titrate the Fe2+ in the dissolved sample to a pink endpoint. How many moles of Fe2+ were in the the sample? How many grams of irion were present in the sample and what was the mass percent iron in the sample?
- 25.0mL of a 0.515 M K2S solution is mixed with 30.0 mL of 0.833 M HNO3 acid solution to give the following reaction: K2S(aq) + 2HNO3(aq) → 2KNO3(aq) + H2S(g) H2S is an unwanted by-product in a pulp and paper industry. To capture H2S gas, it is bubbled through a NaOH solution to produce Na2S with a yield of 94%. H2S(g) + 2NaOH (aq) → Na2S (aq) + 2H2O(l) The mass of H2S(g) that was processed (in kg) if 10.76 kg of Na2S was collected isThe salt content was extracted from a 12.0000 g junk food sample. The extract was diluted to 100.00 mL. From this solution, 15.00 mL was taken and required 23.75 mL of a 0.08943 M AgNO3 solution to reach the endpoint. What is the percentage by mass of salt as NaCl (58.45 g/mol) in the junk food sample?How many grams of Ag2CO3 will be soluble in 250 ml of 0.180 M AgNO3? (Ksp of Ag2CO3 = 5.95 x 10-10) Cr2O72- + 14 H+ + 6 e ----- 2 Cr3+ + 7H2O Eo = - 1.25 VMnO4- + 8 H+ + 5 e ---- Mn2+ + 4 H2O Eo = - 1.27 VFe3+ + e ------ Fe2+ Eo = - 1.20 VZn2+ + 2e --------- Zn Eo = - 0.76 V ( R = 0.08205 L.atm/mol.K; R = 8.314 J/mol.K; F = 96481 C/mol electron) ΔG = ΔH – TΔS ; E = Eo - 0.0257 ln Q ; ΔG = ΔGo + RT ln Q