A face centered cubic cell has an edge length of 495 pm. What is that atomic radius?
1 answer:
Answer:
r = ¼a√2
Explanation:
For a face centered cubic cell, which we call FCC, the atoms at the face diagonals touch each other.
Now, If we denote the edge length of cube to be "a" and the radius of the atom is denoted as "r", as seen in the attached FCC cell:
(√2)a = 4r
Divide both sides by 4 to get;
r = ¼a√2
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Answer:
39.2 g
Explanation:
- 2Ni₂O₃(s) ⟶ 4Ni(s) + 3O₂(g)
First we <u>convert 55.3 grams of Ni₂O₃ into moles of Ni₂O₃</u>, using its<em> molar mass</em>:
- 55.3 g ÷ 165.39 g/mol = 0.334 mol Ni₂O₃
Then we <u>convert 0.334 moles of Ni₂O₃ into moles of Ni</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 0.334 mol Ni₂O₃ *
= 0.668 mol Ni
Finally we <u>calculate how much do 0.668 Ni moles weigh</u>, using the<em> molar mass of Ni </em>:
- 0.668 mol Ni * 58.69 g/mol = 39.2 g