Answer:
Valency of the two elements
Explanation:
In order to correctly formularize the compound, the student needs to ask of the valences of the elements involved.
Chlorine has valency of -1, while Barium +2.
The exchange of valency gives the formula
.
Another explanation is that Barium donates 2 electrons which the chlorine atoms accept, in order to form the covalent bond.
I hope this was helpful.
Oxidizing acids, such as nitric acid, can explode when mixed with what type of incompatible chemicals like alcohol, or charcoal.
An oxidizing acid is a Brønsted acid that is a strong oxidizing agent. most Brønsted acids can act as oxidizing agents because the acidic proton can be reduced to hydrogen gas. a few acids include other structures that act as stronger oxidizing agents than hydrogen ions.
Oxidizing acids include concentrated perchloric acid, nitric acid, iodic acid, chromic acid, and the glass cleansing mixture of chromium trioxide and sulfuric acid.
learn more about acids here brainly.com/question/25148363
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Answer:
Activation energy of phenylalanine-proline peptide is 66 kJ/mol.
Explanation:
According to Arrhenius equation-
, where k is rate constant, A is pre-exponential factor,
is activation energy, R is gas constant and T is temperature in kelvin scale.
As A is identical for both peptide therefore-
![\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7Bala-pro%7D%7D%7Bk_%7Bphe-pro%7D%7D%3De%5E%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-E_%7Ba%7D%5E%7Bala-pro%7D%5D%7D%7BRT%7D)
Here
, T = 298 K , R = 8.314 J/(mol.K) and 
So, ![\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.05%7D%7B0.005%7D%3De%5E%7B%5Cfrac%7B%5BE_%7Ba%7D%5E%7Bphe-pro%7D-%2860000J%2Fmol%29%5D%7D%7B8.314J.mol%5E%7B-1%7D.K%5E%7B-1%7D%5Ctimes%20298K%7D%7D)
(rounded off to two significant digit)
So, activation energy of phenylalanine-proline peptide is 66 kJ/mol
Answer:
6400 molecules / cm^3
Explanation:
10.6*10^-16 mol/L * 6.022*10^23 molecules/mol * 1 L / 1000 cm^3 * 1 / 100 L = 6400 molecules / cm^3
The scientists responsible for testing the hypothesis are already famous