3. Consider the uranium atom, U. a.) (1 pt) Write the electron configuration of

Question

3. Consider the uranium atom, U. a.) (1 pt) Write the electron configuration of Ut for the 23MU isotope, and draw an orbital box diagram of the valence electrons (f orbital electrons are treated like d orbital electrons in v b.)(2 pt) The 23"U isotope is the most abundant in nature, at 99.274%, with the next most common isotope being 235U at 072096. Let's consider a fictional isotopic distribution of U for this problem. In a fictitious universe, U has 3 common isotopes, 2snd4U. Find the abundances of the lighter two isotopes, given that the average isotopic mass for U in this fictitious universe is 237.21 amu, and the abundance of 238U is 77.00%.

Explanation / Answer

b)

(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element

Given

Average atomic weight of U= 237.21

Abundance of U238 = 77%

234 x + 235 y + 238 x (77/100) = 237.21

and

x + y + (77/100) = 1.00

y = 1.00 - (77/100) - x

substitute this on the above equation

234 x + 235 (1.00 - (77/100) - x) + 238 x (77/100) = 237.21

234x + 235 - 180.95 - 235x + 183.26 = 237.21

237.31-x = 237.21

x = 237.31-237.21

x = 0.1

0.1 + y + 0.77 = 1.00

y = 1.00-0.87

y = 0.13

Percentage abudance of U234 is 10%

Percentage abundance of U235 is 13%

c)

The balanced equation for combution of uranium is

U + O2 -----> UO2

Given

density of UO2 is 5.53 g/cm3

Temperature is 32.0oC

Pressure is 1.17 atm

Using ideal gas equation

PV = nRT

Volume of a ring of rectangular cross section = pi * width / 4 (outer_diameter² - inner_diameter²)

Circumference of the planet = 3,263 Miles

C = pi x d

d = C / pi

Diameter of the planet = 3,263 Miles / 3.14 = 1038.6 Miles or 1671.5 km

So the inner diameter of the Ring is 1671.5 + 2(Distance from the earth)

= 1671.5 km + 2(12.9 km)

d = 1697.3 km

Outer diameter of the Ring is 1697.3 + 2(thickness of the ring)

= 1697.3 + 2(2.5700)

D = 1702.44 km

Width of th ring c = 2.420 km

Substituting the on the above equation

= pi * c / 4 (D² - d²)

= 3.14 x 2.420 / 4 (1702.442 - 1697.32 )

Volume of a ring of rectangular cross section 33213.47 km3

PV = nRT

P = 1.17 atm

T = 32.0 oC or 305.15 K

V = 33213.47 km3

R = 8.205 746 × 105 m3 atm K1 mol1

n = 1.17 atm x 3.321347 × 1013 m3 / 8.205 746 × 105 m3 atm K1 mol1 x 305.15 K

n = 1.5519 x 1015 Moles

We know from the ideal gas law that 1 mole of any gas occupy 22.4 L of space.

So,

= 1.5519 x 1015 Moles x 22.4 L / Moles

= 3.476 x 1016 L

In terms of teraliter is 34760

So we need 34760 teraliters of oxygen

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