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The figure shows the bonding of the cytosine and guanine molecules. The O – H and H – N distances are each 0.110nm. In this case, assume that the bonding isdue only to the forces along the O – H – O, N – H – N and O – H – N combinations, and assume also that these three combinations are parallel to each other.

** Calculate the net force** that cytosine exerts on guanine due to the preceding three combinations.

*F*) between two charges ( q1,q2) separated by the distance

*d*is determined by

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k=8.99×10^{9} Nm^{2}/C^{2}

Here is my answer

^{2})/(r^{2}). K=9 x 10^{9}, and e = 1.6 x 10^{-19}. Calculating both correctly will give you 6.38 x 10^{-9}& 1.9 x 10^{-8}.Add the two numbers together and you will get the answer! 1.26 x 10

^{-8}!!!! !It’s mine! First, realize that both the top- and bottom forces cancel each other. If you recall that each atom is assigned a +e/-e, then you can see that they cancel because they are the same length. The only thing you need to be concerned about is the middle one, with a distance of.300nm. The distance between the two bonds N-H is calculated from this. One is ….11 and the other is.19. For each bond, you will use F=k (e2)/(r2). K=9*109, and e = 1.6*10-19. Calculating both correctly will give you 6.38*10-9 & 1.9*10-8. Add the two numbers together and you will get the answer! 1.26*10^-8!!!! !