As Density = Mass/Volume
Mass = 26.0g
Density = 1.44g/mL
Therefore Volume = Mass/Density
=> Volume = 26.0/1.44 = 18.055... = 18.1mL (to 3 sig figs)
Answer:
suspension
Explanation:
a mixture in which particles can be seen and easily separated by settling or filtration.
hope this helps
Mass of Sulphur dioxide : 256 g
<h3>Further explanation</h3>
Given
Reaction
S + O2 --> SO2 *
Required
Mass of Sulphur dioxide
Solution
mol of Sulphur (Ar=32 g/mol) :
mol = mass : Ar
mol = 128 : 32
mol = 4
From the equation, mol ratio S : SO2 = 1 : 1, so mol SO2 = 4
Mass of SO2 :
mass = mol x MW SO2
mass = 4 x 64
mass = 256 g
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.