Par for the Course V
By Chris Morse

Yes, this is the obligatory chemistry puzzle, but all featuring simple
calculations that one can learn in 5.111 or its other incarnations.

Each of the 4 problems listed has one unknown element present.
Solving for each, and taking the symbol for each element in order
gives:

Re-Se-W-N

Problem 1:

With the excess phosphate ions, all three of the salts precipitate:

6.5 mmol of copper(I) ions -> 0.61882 g of copper(I) phosphate
3.5 mmol of iron(III) ions -> 0.527857 g of iron(III) phosphate
5.625 mmol of metal(II) ions

This leaves 1.40332 g left for the metal(II) phosphate salt.  Solving
for the MW of the unknown metal gives 186.17 g/mol which is Rhenium.

Problem 2:

This is a standard combustion question.  You need its weight so that
you can use Graham's Law of Effusion to find the MW of the gaseous
dichloride.

There are 0.272440 moles of H and 0.15892 moles of C, and using the
conservation of mass you can find there must be 1.8166 g of O, or
.113544 moles of O.  This yields an empirical (and molecular) formula
of C7H12O5, which has a weight of 176.17 g/mol.

Graham's Law states:

Rate A/Rate B = square root (MW B/MW A)

Substituting in the times and the masses, yields 149.65 g/mol for the
dichloride.  Subtracting two 35.453's for the Cl's gives, 78.74 g/mol
which is Selenium.

Problem 3:

5 MnO42- +  3 Mn2+ -> 3 MnO4- + 5 M3+

This is the balanced reaction of the titration that is taking place.
The concentration of the manganese two solution is 0.0079679 M.  Which
means that 0.0034036 moles of MO4-2 reacted with it.  So, one gram of
it must have had a molecular weight of 293.81 g/mol.  Removing the
Na's and the O's leaves 183.83 g/mol for the element, which makes it
Tungsten.

Problem 4:

A Bohr atom is a one-electron atom that follows the Bohr model
equation for changing energy states.

E = -Rh Z^2( 1/nf^2 Ð 1/ni^2)

Kinetic Energy = h * nu Ð phi (for the photoelectric effect)

Kinetic Energy = 1/2*m*v^2

1/2*(9.287x105 m/s)^2(9.1095x10^-31 kg) = E Ð 5.65 eV * 1.60219 x 10^-19 J/eV
Which means the photon that hit the metal had an energy of 1.298077 x 10^-18 J.

1.298077 x 10^-18 J = 2.1799 x 10^-18 J * Z^2 (1/36 Ð 1/64)

Solve for Z:  Z = 7 which means that the Bohr atom was Nitrogen (6+)