Rhodium has a face-centered cubic structure and has a density of 12.4 g/cm^3. Wh

Question

Rhodium has a face-centered cubic structure and has a density of 12.4 g/cm^3. What is its atomic radius? 134 pm 1070 pm 268 pm 380 pm a4.55Lof water contains 0.115 g of sodium ions determine the concentration of sodium ions in ppm if the density of the solution is 1.00g/ml 52.3ppm 12.7ppm 13.2ppm25.3ppm 36.5ppm Given the following balanced equation, determine the rate of reaction with respect to[NOCl] 2 NO(g) + Cl2(g) rightarrow 2 NOCl(g) Rate = -2delta[NOCL]/delta t)Rate = -1/2delta[NO]/delta t Rate = -1/2delta[NOCL]/DELTAt Rate = +1/2delta[NOCL]/deltat It is not possible to determine without more information. consider the following reaction at equilibrium what effect will reducing the volume of the reaction mixture have on the system?

Explanation / Answer

8) option A) 134 pm is the answer.

The atomic weight of Rh = 102.9 g mol^-1.

Rh adopts an fcc (ccp) lattice.
There are 4 Rh atoms inside the unit cell of Rh: 8× on corners and 6×½ on faces.
Mass inside cell = 4×AtWt Rh/A where A = Avogadro constant = 6.022×10^23 mol^-1
Mass = 4×102.9/(6.022×10^23) = 6.835×10-22 g
Let side of cell = a; Volume of cell is therefore a^3
D = M/V a^3 = 6.835×10-22/12.4 = 5.512×10^-23 cm^3
a = (5.512×10^-23) = 3.806×10^-8 cm (note density is in cm^-3 so a is in cm).
Important part: the atoms along an edge of the cell do not touch, but atoms along the face diagonal do. From Pythagoras the face diagonal = 2a which it turn = 4rRh where rRh is the radius of a rhodium atom.
rRh = [(2)×3.806×10^-8)]/4 = 1.35×10-8 cm = 1.35 Å (1 Å = 1×10-8 cm) or 135 pm
Lit: Atomic radius (empirical): 135 pm

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