CO2 and H2O are compounds hope this helped :)
Answer:
0.00011765 M
Explanation:
When a solution is prepared by dilution, the volumes and concentrations are related by:
C1*V1 = C2*V2
Where C1 is the concentration of the solution 1, V1 is the volume of the solution 1, C2 is the concentration of solution 2, and V2 is the volume of solution 2.
The stock solution is the solution 1, and the standard solution, the solution 2, so:
0.0001974*29.80 = C2*50.00
C2 = 0.00011765 M
Answer : The value of equilibrium constant for this reaction at 262.0 K is 
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 262.0 K is
Answer:
See explanation
Explanation:
For a reaction that proceeds by E1 mechanism, the rate determining step involves the formation of the carbocation.
The rate of formation of this carbocation depends only on the concentration of the t-butyl bromide since it is the only specie that enters into the rate equation.
Hence, when the concentration of t-butyl bromide is tripled, the rate of reaction is tripled.
Methanol does not enter into the rate equation hence doubling its concentration does not affect the rate of reaction.
The reaction, as what is depicted in the thermonuclear equation is one of the best example of an endothermic reaction. In addition, the endothermic process revolves around the idea that the system can also absorb the energy from its surroundings, in contrast to the idea of releasing its energy to its environment.
