Answer:
We identify nucleic acid strand orientation on the basis of important chemical functional groups. These are the <u>phosphate</u> group attached to the 5' carbon atom of the sugar portion of a nucleotide and the <u>hydroxyl</u> group attached to the <u>3'</u> carbon atom
Explanation:
Nucleic acids are polymers formed by a phosphate group, a sugar (ribose in RNA and deoxyribose in DNA) and a nitrogenous base. In the chain, the phosphate groups are linked to the 5'-carbon and 3'-carbon of the ribose (or deoxyribose) and the nitrogenous base is linked to the 2-carbon. Based on this structure, the nucleic acid chain orientation is identified as the 5'-end (the free phosphate group linked to 5'-carbon of the sugar) and the 3'-end (the free hydroxyl group in the sugar in 3' position).
Answer:

Explanation:
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Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:
![K=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.
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Answer:
60 J
Explanation:
The law of conservation of energy states that energy is neither created nor destroyed, just converted into different forms. This means the total mechanical energy of the object at point A will be the same as the total mechanical energy at point B, and the question tells us the total of that mechanical energy is 150 J. Note we are assuming no energy is lost from the system as heat.
At point B, if the potential energy is 90 J, the remainder of the 150 J total must be kinetic energy. KE = 150 J - 90 J = 60 J.
Answer: The equilibrium constant for the overall reaction is 
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
a) 
![K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
b) 
![K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
For overall reaction on adding a and b we get c
c) 
![K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5E%5Cfrac%7B5%7D%7B2%7D%7D)
![K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_c%3DK_a%5Ctimes%20K_b%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%5Ctimes%20%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
The equilibrium constant for the overall reaction is 