A3.00- g sample of an alloy (containing only pb and sn) was dissolved in nitric acid (hno3). sulfuric acid was added to this solution, which precipitated 1.69 g of pbso4. assuming that all of the lead was precipitated, what is the percentage of sn in the sample?

The reaction between Pb and H2SO4 will be

Pb+2  + SO4^-2 = PbSO4(s)

So each mole of Pb will give one mole of PbSO4

Moles of PbSO4 obtained = mass / molar mass = 1.69 / 303.26 = 0.0056 moles

Moles of Pb present = 0.0056

Mass of Pb present = moles X atomic mass = 0.0056 X 207 = 1.16 grams

So mass of Sn in sample = 3 - 1.16 = 1.84

% of Sn = Mass of Sn X 100 / total mass = 1.84 X 100 / 3 = 61. 33%

i think its 100.

hope this !

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5  m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.