Answer:
i think that the answer is solid liquid and gas.
Explanation:
590 mL = 590 cm³= 0,59 dm³
C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________
M KNO₃ = 39g+14g+16g×3 = 101 g/mol
1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃
:)
Use the state equation for ideal gases: pV = nRT
Data:
V = 88.89 liter
n = 17 mol
T = 67 + 273.15 = 340.15 K
R = 0.0821 atm * liter / (K*mol)
=> p = nRT / V = 17 mol * 0.0821 (atm*liter / K*mol) * 340.15 K / 88.89 liter
p = 5.34 atm
Answer: p = 5.34 atm
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)