The enthalpy of combustion of propane is -2202.0 kJ/mol.
The enthalpy of combustion is released when one mole of compound is burned in the presence of excess oxygen. The bonds between carbon and hydrogen in the compound are broken and bonds between carbon and oxygen are formed as well as bonds between hydrogen and oxygen are formed.
Answer:
The last one is first, the first one is second, and the second one is third.
Explanation:
Answer: Option (d) is the correct answer.
Explanation:
The given equations as as follows.
(slow)
Therefore, overall reaction equation will be as follows.
So, cancelling the spectator ions then the equation will be as follows.
As, it is known that slow step of a reaction is the rate determining step. Therefore, rate law for the slow step will be as follows.
Rate law = ![k[H_{2}O_{2}][I^{-}]](https://tex.z-dn.net/?f=k%5BH_%7B2%7DO_%7B2%7D%5D%5BI%5E%7B-%7D%5D)
Hence, the reaction is first order with respect to ![[I^{-}]](https://tex.z-dn.net/?f=%5BI%5E%7B-%7D%5D)
and it is also first order reaction with respect to ![[H_{2}O_{2}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DO_%7B2%7D%5D)
.
Also, acts as a catalyst in the reaction.
Thus, we can conclude that the incorrect statement is 
is a catalyst.
Answer is D - Argon atom
Potassium has atomic number as 19. Hence, number of protons = 19
If the atom is neutral,
number of protons = number of electrons = 19
Hence, the electron configuration of potassium (K) is
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
Potassium ion is formed by losing an electron. Then the number of electrons = 19 - 1 = 18
Electron configuration of K⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶.
This is same as the electron configuration of Ar which has 18 electrons.
