Answer: A more electronegative atom will have more attraction to the electrons in a chemical bond.
Explanation:
An atom that is able to attract electrons or shared pair of electrons more towards itself is called an electronegative atom.
For example, fluorine is the most electronegative atom.
Due to its high electronegativity it is able to attract an electropositive atom like H towards itself. As a result, both fluorine and hydrogen will acquire stability by sharing of electrons.
Thus, we can conclude that a more electronegative atom will have more attraction to the electrons in a chemical bond.
Answer:
The correct answer is "Iron and oxygen act as Fe3+ and O2− ions respectively, forming rust (Fe₂O₃) in the presence of water by the formation of an ionic bond".
Explanation:
Rust is formed when iron reacts with oxygen in the presence of water (either if the iron is submerged or exposed to moisture in the air), forming the chemical compound Fe₂O₃. The presence of water is needed for rust formation because iron and oxygen act as ions when they are exposed to water, particularly Fe3+ and O2− ions respectively. The bond formed between these two elements are ionic bonds, because it is comprised of the reaction between a metal (iron) and a non-metal (oxygen).
Answer:
If 1000. mL of water freezes, which of the following is a reasonable approximation for the volume of the resulting ice?
Group of answer choices
1000. mL
961 mL
1040 mL
Explanation:
Ice is fewer denser than water.
The reason is the volume occupied by the same mass of ice with water is more than the volume occupied by water. Ice has more empty space within it.
Due to this reason, ice floats on water.
When 1000ml of water freezes to ice then its volume is greater than water.
Among the given options the correct answer is 1040 mL .
Answer:
(a)
(b)
Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:
Best regards.
