An element crystilizers in bcc unit cell with edge length of 290 pm. What is the radius of an atom of element?
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Answers
Atoms cannot be created or destroyed. Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form compound atoms. Atoms of same element can combine in more than
one ratio to form two or more compounds.
REAL ANSWERThe expression for the density of the unit cell is given below:
The expression for the density of the unit cell is given below:d=a3×NAn×M
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,a=288 pm =288×10−10 cm
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,a=288 pm =288×10−10 cmd=7.2 g cm−3
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,a=288 pm =288×10−10 cmd=7.2 g cm−3n=2
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,a=288 pm =288×10−10 cmd=7.2 g cm−3n=2Substituting values in the above expression, we get
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,a=288 pm =288×10−10 cmd=7.2 g cm−3n=2Substituting values in the above expression, we get7.2=(288×10−10)3×6.023×10232×M
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,a=288 pm =288×10−10 cmd=7.2 g cm−3n=2Substituting values in the above expression, we get7.2=(288×10−10)3×6.023×10232×M ⟹M=27.2×(288×10−10)3×6.023×1023=51.8
The expression for the density of the unit cell is given below:d=a3×NAn×MGiven that,a=288 pm =288×10−10 cmd=7.2 g cm−3n=2Substituting values in the above expression, we get7.2=(288×10−10)3×6.023×10232×M ⟹M=27.2×(288×10−10)3×6.023×1023=51.8Hence, the atomic mass of the element is 51.8 g/mol.
Explanation:
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