A public water supply was found to contain 0.8 partper billion (ppb) by mass of chloroform, .(a) How many molecules would be present in a350 mL glass of this water? (b) If the in part (a)could be isolated, would this quantity be detectable onan ordinary analytical balance that measures masswith a precision of g?

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A: Given, Mass of Mg-Al alloy = 1.00 g Mass of H2 produced = 0.107 g   Moles of H2 produced can be…

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A: Given that : Volume of Pb(ClO3)2 = 1.50 L Volume of NaI = 0.900 L Concentration of NaI = 0.130 M…

Q: 3.128 Heating 2.40 g of the oxide of metal X (molar mass ofX = 55.9 g/mol) in carbon monoxide (CO)…

A: The moles of X produced is, number of moles=given massmolar mass…

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A: To find: The difference between electrolyte and non electrolyte

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Q: A 0.2417g sample of a compound composed of C,H,O,Cl only, is burned in oxygen yielding 0.4964g of…

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Q: cer your answer in the provided box. ample of 0.6760 g of an unknown compound containing barium ions…

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A: Given that: Mass of Na2 SO3 = 11.7 g Mass of water = 443 g  Total mass of solution = mass of water +…

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A: As per the question, Mass of = 35 g Let mass of NaCl = x g Then, mass of =(35-x) g  Let the number…

Q: vities [References) TUTOR Molarity: Mass Solute Calculate the mass, in grams, of KNO, required to…

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Q: Suppose 7.00 g of AGNO3 is added to a solution which contains an excess of dissolved KBr. If 7.32g…

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Q: The molar mass of the analyte is known, but not grams in the sample. If 0.9834 grams of sample…

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A: The balanced reaction is given as: CuSO4+Na2CO3→CuCO3+Na2SO4

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Q: How many grams of precipitate is expected for the reaction of Silver nitrate reagent with 0.1000…

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Q: How many moles of Ba(OH)2 are present in 275 mL of 0.600 mol L-1 Ba(OH)2? Express your answer…

A: Volume of Ba(OH) 2= 275 ml = 0.275 L Molarity of Ba(OH) 2 = 0.600 mol L —1

Q: A. A water soluble sample containing an unknown amount of sulfate ion weighed 0.4892g. When treated…

A: Question (A) is answered as per asked.

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A: To calculate the mass of MgI2: First we have to calculate mol of KI,

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A: Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and…

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A: Precipitation reactions: These are the example of double displacement reaction. In these reactions,…

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Q: What is the molarity of the solution when 0.031 moles of NaOH are contained in 5.97 mL of solution?…

A: Welcome to bartleby ! Number of moles = 0.031 moles  Volume = 5.97 ml or 5.97 ×10-3 L

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Q: A sample weighing 3.056 g is a mixture of Fe, O, (molar mass = 159.69 g/mol) and Al, O, (molar mass…

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Q: If 100.0 mL of 0.271 M Na, SO, are added to 100.0 mL of 0.905 M Pb(NO3),, how many grams of PbSO,…

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

A: First we find the moles of PbCl2 , from the moles of PbCl2 and using the Balanced chemical equation…

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Q: Consider the reaction of 30.0 mL of 0.244 M Bal, with 20.0 mL of 0.315 M Na3PO4. What quantity in…

A: 1 M BaI2 solution means 1 mole BaI2 is present in 1000 mL solution. Hence, 0.244 M BaI2 solution…

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A: Given : Concentration of Cr(NO3)3 = 1.75 M Volume of Cr(NO3)3 solution = 150.0 mL = 0.150 L…

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A: Given that -  Volume of oxalic acid dihydrate Solution = 21.49 mL  Mass of calcium oxide = 0.2203 g…

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A: Given, question belong to the Limiting reagent.  The limiting reagent (or limiting reactant or…

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Q: mass, in when 35,5 ml is formed reacted with grams, of Ag of O.184 M AgnNOz O.1824 M AgNOz is and…

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A: 1 M Na2SO3 solution means 1 mole Na2SO3 is present in 1 L means 1000 mL solution.  Molecular weight…

Q: What volume of 0.205 M K, PO, solution is necessary to completely react with 138 mL of 0.0104 M NiC,…

A: K3PO4  molarity = 0.205 M  Volume of K3PO4 = V litre  Number of moles = 0.205 V NiCl2  Molarity =…

Q: 9. In an experiment, a student must make 75.00 mls of 1.400M (NH,),Cro,. How many grams of ammonium…

A: Molarity of (NH4)2CrO4 = 1.400 M = 1.400 mol/L Volume of Solution = 75.00 mL = 0.075 L   We know , *…

Q: without punctuation, e.g, ABC.) Without doing any calculations it is possible to determine…

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