A portion of the periodic table is shown below: A portion of the thirteenth and fourteenth columns of the periodic table is show
2 answers:
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Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:
To find,
Using 1 and 2 , we get:
<u>Size of the potassium ion = 1.33 Å</u>
Answer:
Explanation:
Given parameters:
Mass of CuCl₂ = 2.50g
Mass of Al = 0.50g
Unknown:
Number of moles of CuCl₂ and Al = ?
Solution:
To solve this problem, we must understand that the number of moles is a fundamental property used in stoichiometry calculations.
Number of moles =
Molar mass of CuCl₂ = 63.6 + 2(35.5) = 134.5g/mole
Molar mass of Al = 26.98g/mole
Number of moles of CuCl₂ = = 0.019moles
Number of moles of Al = = 0.019moles