Multiple Choice
Using the phase diagram below, determine the melting point of an alloy that is 60% nickel.
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24. Transition Metals and Coordination Compounds
Atomic Radius & Density of Transition Metals
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Determine the diameter of the octahedral hole in the interstitial alloy titanium carbide. The atomic radius of titanium is 147 pm.
A
30.5 pm
B
60.9 pm
C
121.7 pm
D
355.1 pm
E
710.2 pm
Verified step by step guidance1
Understand that in an octahedral hole, the atoms are arranged in such a way that the hole is surrounded by six atoms. The size of the hole is determined by the arrangement and size of these surrounding atoms.
Recognize that the octahedral hole is formed in the crystal lattice of titanium carbide, where titanium atoms are arranged in a close-packed structure, and the carbon atoms occupy the octahedral holes.
The diameter of the octahedral hole can be calculated using the formula for the radius of the octahedral hole, which is approximately 0.414 times the radius of the surrounding atoms. In this case, the surrounding atoms are titanium with a radius of 147 pm.
Calculate the radius of the octahedral hole using the formula: \( r_{hole} = 0.414 \times r_{Ti} \), where \( r_{Ti} \) is the radius of titanium.
Double the radius of the octahedral hole to find the diameter: \( d_{hole} = 2 \times r_{hole} \). This will give you the diameter of the octahedral hole in the titanium carbide structure.
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