Homework Answers
1.
Trial (1)
Mass of AgCl = 0.387 g.
Molar mass of AgCl = 107.87 g/mol
Amount of AgCl = mass / molar mass = 0.387 / 107.87 = 0.00359 mol
From the fromula,
1 mol AgCl has 1 mol Cl
then, 0.00359 mol AgCl has 0.00359 mol Cl
Therefore,
Mass of Cl = moles * molar mass = 0.00359 * 35.45 = 0.127 g.
2.
Trial (1)
% by mass of Cl = Mass of Cl * 100 / Mass of unknown sample = 0.127 * 100 / 0.354 = 35.9 %
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for a lab observing Boyle’s Law.
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and HSO-3 and rate
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SAFETY PRECAUTIONS BEYOND STANDARD PROCEDURES CAUTION! Potassium chlorate is a strong oxidizing agent and may cause fires or explosions if mixed or heated with combustible materials. Observe the following safety precautions when working with potassium chlorate. Heat samples slowly and carefully to avoid spills and spattering. Check before disposing of potassium chlorate residue. Dispose of solutions containing...
Need help graphing these points, finding the slope and percent
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I need help finding how to do the calculations for these
tables. If you could show me how to do the 1st determination/ trial
1 for each of them, I can figure out how to do the second and third
determinations/trials. If you could also show me how to do the
standard deviations, that'd be much appreciated. Thank you!
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I just need help with question
D, finding the ksp value.
AgCI (s) = Ag+ (aq) + Cl^- (aq) lg the data in the table given: Calculate the DeltaHdegree (in kJ/mol) for dissolving AgCI in aqueous solution. Calculate the ASdegree (in J/molK) for dissolving AgCI in aqueous solution. Calculate the AGdegree (in kJ/mol) for dissolving AgCI in aqueous solution at 25degreeC. What is the value for Ksp at this temperature?
hey i really need help for the molality to mass question. the
solvent is water and 100g. SOS!
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