acs gchem question. page 39

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ddsshin

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I am not sure aboutt question 19 which asks:
which sets are the substances arranged in order of decreasing solubility in water. Answer is D
which is AgCl > AgBr > AgI
Can anyone explain this??????
and another question. this is from page 37 #8
copper crystalizes in a face-centered cubic lattice. if edge of unit cell is 351 pm. what is radius of copper atom? Answer is 128pm. but I have no idea how to get this...
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Ag-Cl= (+)------(-) the minus in this bond (Cl) is more negative.
Ag-Cl bond has a larger polarity than the other two. thus it dissolves more in a polar substance. more polar=more soluble in a polar substance. Cl is more electronegative than I.
 
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this was from p39 Qs25
the edge of body centered cubic unit cell of element was found to be 3.16X10^-8 cm. density of metal is 19.35g/cm^3. what is molar mass???
 
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ddsshin said:
this was from p39 Qs25
the edge of body centered cubic unit cell of element was found to be 3.16X10^-8 cm. density of metal is 19.35g/cm^3. what is molar mass???
Click to expand...

You're given the edge. So, you can calculate the volume of the unit cell, which is in shape of a cube. Therefore:

To simplify calculations, let's assume:
Edge = 3x10^-8cm

V = (edge)^3 = (3x10-8)^3 = 27x10^-24 cm^3

Now that you have the volume and density you can go ahead and calculate the mass via the following formula:

Density = mass/volume

So:

mass = Density x volume = (19.35)(27x10^-24)
 
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ddsshin said:
I am not sure aboutt question 19 which asks:
which sets are the substances arranged in order of decreasing solubility in water. Answer is D
which is AgCl > AgBr > AgI
Can anyone explain this??????
and another question. this is from page 37 #8
copper crystalizes in a face-centered cubic lattice. if edge of unit cell is 351 pm. what is radius of copper atom? Answer is 128pm. but I have no idea how to get this...
Click to expand...

Consider the following figure:
123%2011%20face%20centered%20cell%20p.464.jpg


All you need to do now is to use the Pythagorean Theorm. So:

(Edge)^2 + (Edge)^2 = (4r)^2 = 16(r^2)

Now plug in the length of each edge and solve for "r"
This should give you the correct answer for the radius of each atom!
 
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