Answer:
The particles in the substance become less active
Answer:
Sn₃(PO₄)₄ - tin(IV) phosphate.
Explanation:
Hope it helps! :)
Matter buddy, ur welcome :)
Answer: The concentrations of
at equilibrium is 0.023 M
Explanation:
Moles of
= 
Volume of solution = 1 L
Initial concentration of
= 
The given balanced equilibrium reaction is,

Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :

By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of
at equilibrium is 0.023 M
The rate law equation for Ozone reaction
r=k[O][O₂]
<h3>Further e
xplanation</h3>
Given
Reaction of Ozone :.
O(g) + O2(g) → O3(g)
Required
the rate law equation
Solution
The rate law is a chemical equation that shows the relationship between reaction rate and the concentration / pressure of the reactants
For reaction
aA + bB ⇒ C + D
The rate law can be formulated:
![\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cboxed%7B%5Cbold%7Br~%3D~k.%5BA%5D%5Ea%5BB%5D%5Eb%7D%7D%7D)
where
r = reaction rate, M / s
k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹
a = reaction order to A
b = reaction order to B
[A] = [B] = concentration of substances
So for Ozone reaction, the rate law (first orde for both O and O₂) :
![\tt \boxed{\bold{r=k[O][O_2]}}](https://tex.z-dn.net/?f=%5Ctt%20%5Cboxed%7B%5Cbold%7Br%3Dk%5BO%5D%5BO_2%5D%7D%7D)