Answer:
[CO] = 7.61x10⁻³M
7.61x10⁻³x10³ = 7.61
Explanation:
For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:
![Kc = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BC%5D%5Ecx%5BD%5D%5Ed%7D%7B%5BA%5D%5Eax%5BB%5D%5Eb%7D)
We need to know the molar concentrations in the equilibrium. In the beginning, there is only COCl₂, and its concentration is the number of moles divided by the volume:
[COCl₂] = 7.73/10.0 = 0.773 M
So, the equilibrium will be:
COCl₂(g) ⇆ CO(g) + Cl₂(g)
0.773 0 0 <em>Initial</em>
-x +x +x <em> Reacts</em>
0.773-x x x <em>Equilibrium</em>
Supposing that x<<0.773, then:
7.5x10⁻⁵ = x²/0.773
x² = 5.7975x10⁻⁵
x = √5.7975x10⁻⁵
x = 7.61x10⁻³ M
The supposing is correct, so [CO] = 7.61x10⁻³ x 10³ = 7.61
Answer by Mimiwhatsup: An amplitude is a node and an point along a standing wave where the wave has minimum amplitude.
1.) scale of the chart
2.) Notes of the chart
3.) chart symbols
4.) chart corrections
5.) GPS positions
6.) radar fixes
7.) Visual fixtures and position circle and position line
Answer:
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Explanation:
According to the Arrhenius equation,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate of reaction at 
= rate of reaction at 
= activation energy of the reaction
R = gas constant = 8.314 J/K mol
![\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B300%2C000%20J%2Fmol%7D%7B2.303%5Ctimes%208.314%20J%2FK%20mol%7D%5B%5Cfrac%7B1%7D%7B798.15%20K%7D-%5Cfrac%7B1%7D%7B898.15%20K%7D%5D)
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Oxidation number is charge of element in compound. Can be neutral, positive or negative.
Oxygen in dichromate has oxidation number -2, becauce there are seven oxygens, net oxidation number of oxygen is -14.
2·x(oxidation number of Cr) + 7· (-2) = -2 2x= +12
x= +6, oxidation number of one chromium is +6.
