Hi!
The correct option would be 3.85x10^(24)
To find the number of atoms in 250g of potassium, we need to first calculate the number of atoms in
1 mole of Potassium = 39g which contains 6.022x10^(23) atoms of K
<em>(Avogadro's constant value for the amount of molecules/atoms in one mole of any substance)</em>
<em>Solution</em>
So as 39g of Potassium contains 6.022x10^(23) K atoms
1g of Potassium would contain 6.022x10^(23) / 39 = 1.544 x10^(22) atoms
So 250g of Potassium would contain 1.544x10^(22) x 250 = 3.86x10^(24) atoms
The net ionic equation of the reaction could be determined by cancelling out the like ions between both sides of the reaction. These ions are called spectator ions. They are called as such because they do not actively participate in the reaction. The spectator ions are Na+ and Cl-. When you cancel those, the equation would become letter D.
Hey! The answer would be 1/2.
Answer:
b. Second order in NO and first order in O₂.
Explanation:
A. The mechanism
![\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)](https://tex.z-dn.net/?f=%5Crm%202NO%5Cxrightarrow%5Bk_%7B-1%7D%5D%7Bk_%7B1%7D%7DN_%7B2%7DO_%7B2%7D%20%5C%2C%20%28fast%29%5C%5C%5Crm%20N_%7B2%7DO_%7B2%7D%20%2B%20O_%7B2%7D%5Cxrightarrow%7Bk_%7B2%7D%7D%202NO_%7B2%7D%20%5C%2C%20%28slow%29)
B. The rate expressions
![-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]](https://tex.z-dn.net/?f=-%5Cdfrac%7B%5Ctext%7Bd%5BNO%5D%7D%20%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B1%7D%5B%5Ctext%7BNO%5D%7D%5E%7B2%7D%20-%20k_%7B-1%7D%20%5B%5Ctext%7BN%7D_%7B2%7D%5Ctext%7BO%7D_%7B2%7D%5D%5E%7B2%7D%5C%5C%5C%5C%5Crm%20-%5Cdfrac%7B%5Ctext%7Bd%5BN%24_%7B2%7D%24O%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-%5Cdfrac%7B%5Ctext%7Bd%5BO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B2%7D%5B%20N_%7B2%7DO_%7B2%7D%5D%5BO_%7B2%7D%5D%20-%20k_%7B1%7D%20%5BNO%5D%5E%7B2%7D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BNO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%3D%20k_%7B2%7D%5B%20N_%7B2%7DO_%7B2%7D%5D%5BO_%7B2%7D%5D)
The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.
C. Assume the first step is an equilibrium
If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.
![\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}](https://tex.z-dn.net/?f=%5Crm%20k_%7B1%7D%5BNO%5D%5E%7B2%7D%20%3D%20k_%7B-1%7D%20%5BN_%7B2%7DO_%7B2%7D%5D%5C%5C%5C%5C%5Crm%20%5BN_%7B2%7DO_%7B2%7D%5D%20%3D%20%5Cdfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D)
D. Substitute this concentration into the rate law
![\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B%5Ctext%7Bd%5BNO%24_%7B2%7D%24%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%3D%20%5Cdfrac%7Bk_%7B2%7Dk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%20%5BO_%7B2%7D%5D%20%3D%20k%5BNO%5D%5E%7B2%7D%20%5BO_%7B2%7D%5D)
The reaction is second order in NO and first order in O₂.
Answer : The correct option is, (B) 2, 3 and 6.
Explanation :
Combustion reaction : It is a type of reaction where a hydrocarbon react with an oxygen molecule to give carbon dioxide, water as a product.
In general,

For example : Ethane react with oxygen to give carbon dioxide and water.

In the given list,
is a reactant and
are the products.
Therefore, the correct answer is, (B) 2, 3, and 6