Answer:
Neutron radiation
Explanation:
Neutron radiation is a form of ionizing radiation that presents as free neutrons. Typical phenomena are nuclear fission or nuclear fusion causing the release of free neutrons, which then react with nuclei of other atoms to form new isotopes—which, in turn, may trigger further neutron radiation.
Answer:
a. because of how the earth circulates around the sun.
Answer:
If carbon which has an electrongavity of 2.5 bonds with hydrogen, which has an electrongavity of 2.1, the bond between the two atoms will be classified as a(n) covalent bond → Totally TRUE
Explanation:
We must follow the Pauling's rules to solve these question.
The electonegativity of an element is defined as the ability it has to attract electrons that link it to another element. It is a periodic property, which allows predicting the polarity of the bond formed between two atoms, as well as its covalent or ionic character.
The Pauling scale values range from 0.7 for Francium (the least electronegative element) to 4 for Fluorine, the most electronegative. Using the electronegativities table we can predict the polarity of some compounds
ΔEn (C-H) = 2.5 - 2.1 = 0.4 → This is a covalent bond.
ΔEn ≤ 0.4 → Covalent non polar bond
0.5 < ΔEn < 1.7 → Covalent polar bond
ΔEn ≥ 1.7 → Ionic bond
Answer: The value of ![[H_{3}O^{+}]](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
is 0.0012 M and ![[OH^{-}]](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D)
is 
.
Explanation:
pH is the negative logarithm of concentration of hydrogen ion.
It is given that pH is 2.89. So, the value of concentration of hydrogen ions is calculated as follows.
![pH = - log [H^{+}]\\2.89 = - log [H^{+}]\\conc. H^{+} = 0.0012 M](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5C2.89%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20H%5E%7B%2B%7D%20%3D%200.0012%20M)
The relation between pH and pOH value is as follows.
pH + pOH = 14
0.0012 + pOH = 14
pOH = 14 - 0.0012 = 13.99
Now, pOH is the negative logarithm of concentration of hydroxide ions.
Hence, is calculated as follows.
![pOH = - log [OH^{-}]\\13.99 = - log [OH^{-}]\\conc. OH^{-} = 1.02 \times 10^{-14} M](https://tex.z-dn.net/?f=pOH%20%3D%20-%20log%20%5BOH%5E%7B-%7D%5D%5C%5C13.99%20%3D%20-%20log%20%5BOH%5E%7B-%7D%5D%5C%5Cconc.%20OH%5E%7B-%7D%20%3D%201.02%20%5Ctimes%2010%5E%7B-14%7D%20M)
Thus, we can conclude that the value of is 0.0012 M and is .
Before it is released it as potential energy and after it has been released it transforms into kinetic energy.
