<span>XY4Z2-->Square planar (Electron domain geometry: Octahedral) sp3d2
XY4Z-->Seesaw (Electron domain geometry: Trigonal bipyramidal) sp3d
XY5Z-->Square pyramidal (Electron domain geometry: Octahedral) sp3d2
XY2Z3-->Linear (Electron domain geometry: Trigonal bipyramidal) sp3d
XY2Z-->Bent (Electron domain geometry: Trigonal planar) sp2
XY3Z-->Trigonal pyramidal (Electron domain geometry: Tetrahedral) sp3
XY2Z2-->Linear (Electron domain geometry: Tetrahedral) sp3
XY3Z2-->T shaped (Electron domain geometry: Trigonal bipryamidal) sp3d
XY2-->Linear (Electron domain geometry: Linear) sp
XY3 Trigonal planar (Electron geometry: Trigonal planar) sp2
XY4-->Tetrahedral (Electron domain geometry: tetrahedral) sp3
XY5-->Trigonal bipyramidal (Electron domain geometry: Trigonal bipyramidal) sp3d
XY6-->Octahedral (Electron domain geometry: Octahedral) sp3d2</span>
Answer:
I thinks it Br hope this helps
It would be Atom 2 because the proton and neutron are both nine and the election is 8 which is -1 to nine <span />
Answer : The concentration after 17.0 minutes will be, 
Explanation :
The expression for first order reaction is:
![[C_t]=[C_o]e^{-kt}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%5BC_o%5De%5E%7B-kt%7D)
where,
![[C_t]](https://tex.z-dn.net/?f=%5BC_t%5D)
= concentration at time 't' (final) = ?
![[C_o]](https://tex.z-dn.net/?f=%5BC_o%5D)
= concentration at time '0' (initial) = 0.100 M
k = rate constant = 
t = time = 17.0 min = 1020 s (1 min = 60 s)
Now put all the given values in the above expression, we get:
![[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%280.100%29%5Ctimes%20e%5E%7B-%285.40%5Ctimes%2010%5E%7B-3%7D%29%5Ctimes%20%281020%29%7D)
![[C_t]=4.05\times 10^{-4}M](https://tex.z-dn.net/?f=%5BC_t%5D%3D4.05%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration after 17.0 minutes will be,
Quantity of K2S m = 0.105 m
Number of ions i = 2(K) + 1(S) = 3
Freezing point depression constant of water Kf = 1.86
delta T = i x m x Kf = 3 x 0.105 x 1.86 = 0.586
Freezing point = 0 - 0.586 = 0.586 C
Boiling point constant of water Kb = 0.512
delta T = i x m x Kb = 3 x 0.105 x 0.512 = 0.161
Boiling point = 100 + 0.161 = 100.161 C
