Answer:
See explanation
Explanation:
Many organic compounds have low melting points. This is due to the fact that many of these compounds are non polar.
However, compound X is slightly polar but still has a melting point which is far less than that of sand composed of a high melting point inorganic material.
Since sand has a much higher melting point compared to compound X, the researcher need not be worried that sand was spilled into his beaker.
Explanation:
Because when you move the decimal point once, it it equal to 10 and to convert in the metric system, you must multiply or divide by 10. For example, when converting 4 centimeters to millimeters, you can multiply by 10 to get 40 millimeters or you can move the decimal point to the right and add another 0 to get 40.
Answer:
4.75 is the equilibrium constant for the reaction.
Explanation:

Equilibrium concentration of reactants :
![[CO]=0.0590 M,[H_2O]=0.00600 M](https://tex.z-dn.net/?f=%5BCO%5D%3D0.0590%20M%2C%5BH_2O%5D%3D0.00600%20M)
Equilibrium concentration of products:
![[CO_2]=0.0410 M,[H_2]=0.0410 M](https://tex.z-dn.net/?f=%5BCO_2%5D%3D0.0410%20M%2C%5BH_2%5D%3D0.0410%20M)
The expression of an equilibrium constant is given by :
![K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)


4.75 is the equilibrium constant for the reaction.
Answer:
0!
Explanation:
- You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:
-COOH -NH3
pH 7------------------------------------------------------
Asn - +
Gly - +
Leu - +
- Now that we have our table we'll sketch our peptide's structure:
<em>HN-Asn-Gly-Leu-COOH</em>
This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:
+1 (HN-Asn)
-1 (Leu-COOH)
=0
The net charge is 0!
I hope you find this information useful and interesting! Good luck!