1.)In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yie
ID: 626366 • Letter: 1
Question
1.)In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole product. In a second experiment, the same sulfide was produced in the reaction of 1.58 mercury and 1.10 sulfur what mass of the sulfide of mercury was produced in the second experiment? what mass remained unreacted in the second experiment? express in two sig. figs.
2.)During a severe episode of air pollution the concentration of lead in the air was observed to be 3.07 ug Pb/m^3. How many Pb atoms would be present in a 0.500-L sample of this air (the approximate lung capacity of a human adult)?
3.) how many atoms are present in a 85.0-cm^3 sample of plumbers solder, a lead tin alloy containing 67% Pb by mass and having a density of 9.4 g/cm^3? express in two significant figures.
Explanation / Answer
1. Mercury combines with sulfur as follows - Hg + S = HgS Hg = 200,59 S = 32,066 Therefore 1,54 g of Hg will react with - 1,54 X 32,066 / 200,96 of sulfur. = 0,2462 g S This will form 1,54 + 0,2462 g HgS = 1,7862 g HgS The amount of S remaining = 1,08 - 0,2462 = 0,8338 g 3. D = m/v You are given both density and volume, solve for the mass of the solder. From there, multiply the mass by 67% to get the mass of lead. The remaining 33% of mass is the mass of tin. Convert the mass of lead into moles of lead: (given mass/molar mass = moles) and do the same for tin There are 6.022E23 atoms in 1 mole of any substance. So to find the number of atoms in a substance: 6.022E23 X moles of substance = atoms in substance
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