Depends on the length of the Ant. If it's a millimeter long then the park would be 999,999mm longer than the ant. (I believe)
<span> weather is the condition of the climate in a particular place at a particular
time.</span>
Answer:
I and IV are the correct answers
Explanation:
This reaction has an equilibrium constant of
at 298K.
![CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)](https://tex.z-dn.net/?f=CO%28g%29%2B2H_2%28g%29%5Crightleftharpoons%20CH_3OH%28g%29)
Calculate Kp for each reaction and predict whether reactant or products will be favoured at equilibrium.
A) ![CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)](https://tex.z-dn.net/?f=CH_3OH%28g%29%5Crightleftharpoons%20CO%28g%29%2B2H_2%28g%29)
B) ![\frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DCO%28g%29%2BH_2%28g%29%5Crightleftharpoons%20%5Cfrac%7B1%7D%7B2%7DCH_3OH%28g%29)
C) ![2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g)](https://tex.z-dn.net/?f=2CH_3OH%28g%29%5Crightleftharpoons%202CO%28g%29%2B4H_2%28g%29)
Answer:
A)
:
: reactants are favoured
B)
:
: products are favoured
C)
:
: reactants are favoured
Explanation:
1. We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:
![CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)](https://tex.z-dn.net/?f=CH_3OH%28g%29%5Crightleftharpoons%20CO%28g%29%2B2H_2%28g%29)
![K_{p}'=(\frac{1}{K_p})](https://tex.z-dn.net/?f=K_%7Bp%7D%27%3D%28%5Cfrac%7B1%7D%7BK_p%7D%29)
![K_{p}'=(\frac{1}{2.26\times 10^4})=0.442\times 10^-4}](https://tex.z-dn.net/?f=K_%7Bp%7D%27%3D%28%5Cfrac%7B1%7D%7B2.26%5Ctimes%2010%5E4%7D%29%3D0.442%5Ctimes%2010%5E-4%7D)
As
is less than 1, reactants will be favored at equilibrium.
2. We need to calculate the equilibrium constant for the half equation of above chemical equation, which is:
![\frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DCO%28g%29%2BH_2%28g%29%5Crightleftharpoons%20%5Cfrac%7B1%7D%7B2%7DCH_3OH%28g%29)
![K_{p}'=(\sqrt{K_p})](https://tex.z-dn.net/?f=K_%7Bp%7D%27%3D%28%5Csqrt%7BK_p%7D%29)
![K_{p}'=(\sqrt{2.26\times 10^4})=1.50\times 10^2}](https://tex.z-dn.net/?f=K_%7Bp%7D%27%3D%28%5Csqrt%7B2.26%5Ctimes%2010%5E4%7D%29%3D1.50%5Ctimes%2010%5E2%7D)
As
is greater than 1, products will be favored at equilibrium.
3. We need to calculate the equilibrium constant for the reverse and twice the equation of above chemical equation, which is
![2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g)](https://tex.z-dn.net/?f=2CH_3OH%28g%29%5Crightleftharpoons%202CO%28g%29%2B4H_2%28g%29)
![K_{p}'=(\frac{1}{K_p})^2](https://tex.z-dn.net/?f=K_%7Bp%7D%27%3D%28%5Cfrac%7B1%7D%7BK_p%7D%29%5E2)
![K_{p}'=(\frac{1}{2.26\times 10^4})^2=4.42\times 10^{-9}](https://tex.z-dn.net/?f=K_%7Bp%7D%27%3D%28%5Cfrac%7B1%7D%7B2.26%5Ctimes%2010%5E4%7D%29%5E2%3D4.42%5Ctimes%2010%5E%7B-9%7D)
As
is less than 1, reactants will be favored at equilibrium.
Answer:A precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined. The insoluble salt that falls out of solution is known as the precipitate, hence the reaction's name. Precipitation reactions can help determine the presence of various ions in solution.
Explanation: