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>
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Answer:
The answer to your question is: 4.5 %
Explanation:
0.126 moles of AgNO3
mass percent = ?
mass of water = 475 g
Formula
weight percent = weight of solute / weight of solution x 100
Weight of solute
MW AgNO3 = 108 + 14 + (16 x 3)
= 108 + 14 + 48
= 170 g
170 g of AgNO3 ------------------- 1 mol
x --------------------- 0.126 moles
x = (0.126 x 170) / 1 = 21.42 g of AgNO3
Weight percent = 21.42/475 x 100
= 4.5 %
I can not answer your question because i am a person who is not smart
and will never become smart
Answer:
4.09ₓ10²³ atoms are contained in 10g of C₃H₈
Explanation:
1 mol of C₃H₈ has 3 mol of carbon and 8 mol of hydrogen.
1 mol of C₃H₈ weighs 44 g
In 44 g of C₃H₈ we have 3 mol of C
In 10 g of C₃H₈ we'll have (10 . 3)/ 44 = 0.681 mol
1 mol of C has NA (6.02x10²³) particles
0.681 mol will have (0.681 . 6.02x10²³) = 4.09ₓ10²³ atoms
