The reactions are a bit poorly written. While it's true that aqueous H₂CO₃ is produced in this neutralization reaction, the H₂CO₃ rapidly decomposes to yield CO₂(g) and H₂O(l). Writing the product as H₂CO₃(aq) in the net ionic equation is unnecessarily confusing since it portrays the substance as nonionizing yet water-soluble.
In any case, the Na⁺ and the Cl⁻ are the spectator ions here.
Elements that are in the same columns (vertical) are chemically similar.
Carbon- Si
Cesium- Na
Krypton- Ar
Magnesium- Ca
Al- try Boron
Br try Cl
These are just the ones I picked but it could be any in the same column
Answer:
The rate law for second order unimolecular irreversible reaction is
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Explanation:
A second order unimolecular irreversible reaction is
2A → B
Thus the rate of the reaction is
![v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}](https://tex.z-dn.net/?f=v%20%3D%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20k.%5BA%5D%5E%7B2%7D)
rearranging the ecuation
![-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Integrating between times 0 to <em>t </em>and between the concentrations of
to <em>[A].</em>
![\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_t%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%5Cint%5Climits%5EA_%7B0%7D%20_A%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Solving the integral
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Answer:
They have electrons in their 3d- and 4s-orbital for bond formation.
Explanation:
d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.
The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.
If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.
If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2
it would be A ,inorganic Compound