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

Answer:
RESEWN

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+)