D is the answer. Just look at the picture on your test ❤
Answer
is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) =
Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁
= 0,000643 1/s.<span>
k</span>₂
= 0,00828 1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
<span>
1/T</span>₁ =
1/622 K = 0,0016 1/K.<span>
1/T</span>₂ =
1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol ·
(-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
use the equation q=mcΔt. convert 13 mg to g. so it becomes .013g which will be the mass. C is specific heat of water which is always 4.184. Delta t is 75-1 which is 74. (.013)(4.184)(74) = 4.025
Answer:
The answer to your question is d. 0.5 M
Explanation:
Data
[A] = 1M
K = 0.5
Concentration of B and C at equilibrium = x
Concentration of A at equilibrium = 1 - x
Equation of equilibrium
k = ![\frac{[B][C]}{A}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%5BC%5D%7D%7BA%7D)
Substitution
![0.5 = \frac{[x][x]}{1 - x}](https://tex.z-dn.net/?f=0.5%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B1%20-%20x%7D)
Simplification
0.5 = 
Solve for x
0.5(1 - x) = x²
0.5 - 0.5x = x²
x² + 0.5x - 0.5 = 0
Find the roots x₁ = 0.5 x₂ = -1
There are no negative concentrations so the concentration of A at equilibrium is
[A] = 1 - 0.5
= 0.5 M