Boiling liquids that want to escape gas
Answer:
The activation energy for this reaction = 23 kJ/mol.
Explanation:
Using the expression,

Where,


is the activation energy
R is Gas constant having value = 8.314×10⁻³ kJ / K mol

The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (280 + 273.15) K = 553.15 K
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (376 + 273.15) K = 649.15 K
So,




<u>The activation energy for this reaction = 23 kJ/mol.</u>
Answer:
At equilibrium, the concentration of
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no
nor
present. Immediately,
and
are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: [
]=0 ; [
]= 0 ; [
]=0.60M
C: [
]=+x ; [
]= +x ; [
]=-2x
E: [
]=0+x ; [
]= 0+x ; [
]=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 




At equilibrium, the concentration of
is going to be 0.30M
Answer:
Glucose and oxygen are required for cellular respiration. As the law of conversation states, in a biochemical reaction, mass is conserved. The mass of hydrogen in the glucose is therefore conserved in the water molecules products.