ACS GC question

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LetsSmile

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I have two question from the ACS GC book.

1) A mixture containing 9 mol of F2 and 4 mol of S is allowed to react.

3 F2 + S -> SF6

how many moles of F2 remain after 3 mol of S have reacted?

The answer is 1 but I'm keep getting 0. How do you get 1?


2) copper crystallizes in a face-centered cubic lattice. if the edge of the unit cell is 351 pm, what is the radius of the copper atom?

129 pm



Thank you for your help!!
 
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1) I believe that their answer is wrong. If anyone else knows how to get 1 let us know.

2) On this I get about 125 but which is only 3 off from what the book says. If anyone knows how to get the exact answer let us know but what I did is use the Radius to Side formula. There is one for Simple, Face, and Body centered cubic unit cell.

Simple--> 2r=s

Body Centered --> 4r= S x sqrroot(3)

Face Centered --> 4r= S x squrroot(2)

Hope this helps! If anyone has more input please share.
 
1) I believe that their answer is wrong. If anyone else knows how to get 1 let us know.

2) On this I get about 125 but which is only 3 off from what the book says. If anyone knows how to get the exact answer let us know but what I did is use the Radius to Side formula. There is one for Simple, Face, and Body centered cubic unit cell.

Simple--> 2r=s

Body Centered --> 4r= S x sqrroot(3)

Face Centered --> 4r= S x squrroot(2)

Hope this helps! If anyone has more input please share.


Thank you for the reply! 🙂

Did you get 0 for 1), too?



I got another question.

Use the bond energies in the table to determine H change for the formation of hydrazine, N2H4, from nitrogen and hydrogen according to this equation.

N2 + 2 H2 -> N2H4

Bond Energies, kJ.mol-1:

N-N 159
N=N 201
NN (triple bond) 941
H-H 436
H-N 389

I used the (bonds broken)-(bonds formed) equation and got +56.
But the answer is +98.

I hope there is explanation in the ACS book....T.T
 
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I got another question.

Use the bond energies in the table to determine H change for the formation of hydrazine, N2H4, from nitrogen and hydrogen according to this equation.

N2 + 2 H2 -> N2H4

Bond Energies, kJ.mol-1:

N-N 159
N=N 201
NN (triple bond) 941
H-H 436
H-N 389

I used the (bonds broken)-(bonds formed) equation and got +56.
But the answer is +98.

I hope there is explanation in the ACS book....T.T


Ok, I am not sure what you did but maybe you just miscalculated or something. This is what I got for the equation before doing the math

941+ 2x436 ---> 159 + 4x389

1813-1715= +98

Remember you are breaking a triple bond between the two nitrogens and forming only a single bond because two hydrogens are bonded to each nitrogen. Hope this helps... keep the questions coming

Oh and I also got 0 for your first question
 
Ok, I am not sure what you did but maybe you just miscalculated or something. This is what I got for the equation before doing the math

941+ 2x436 ---> 159 + 4x389

1813-1715= +98

Remember you are breaking a triple bond between the two nitrogens and forming only a single bond because two hydrogens are bonded to each nitrogen. Hope this helps... keep the questions coming

Oh and I also got 0 for your first question


oh..
the two nitrogens form a single bond.. that's why...
I thought they would form a double bond.
Thank you so much! 🙂
 
oh..
the two nitrogens form a single bond.. that's why...
I thought they would form a double bond.
Thank you so much! 🙂


more questions..😀

1) A saturated soln of MgF2 contains 1.6x10-3 mol of MgF2 per liter at a certain temperature. What is the Ksp of MgF2 at this temperature?

The answer is 6.2 x 10-9.

But I got 1.6 x 10-8 because I did...
(1.6x10-3) x ((1.6x10-3)2)^2

What did I do wrong?


2) For the reaction:
NH4Cl -> NH3 + HCl

change of standard enthalpy = +176 KJ
change of standard Gibb free energy = +91.2 KJ
at 298 K

what is the value of Gibb free energy at 1000 K?

the answer is -109KJ

Do you need the change of standar enthalpy?
 
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