Answer: "The reactants are higher in energy than the products"
Explanation:
The exothermic reactions are characterized by the release of heat to the surroundings. The reactants lose heat that is delivered to the surroundings which implies that the products will be lower in energy than the reactants.
The hills that you can see in a reaction energy diagram are not related with the final change of energy. The hills are an indication of the activation energy needed to start the reaction, but they do not measure the change of energy from the products to the reactants.
The enthalpy that is a state variable that identifies the content of heat. Then the change of enthalpy for the exothermic reactions is negative, meaning that the energy of the products is lower than the energy of the reactants.
Answer:
the answer is photosynthesis
Answer:
1.02mol
Explanation:
Using the general gas equation below;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the information provided in this question,
P = 2.0 atm
V = 11.4L
T = 273K
n = ?
Using PV = nRT
n = PV/RT
n = 2 × 11.4/ 0.0821 × 273
n = 22.8/22.41
n = 1.017
n = 1.02mol
This is an incomplete question, here is a complete question.
Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.
CaCO₃, Ksp = 8.7 × 10⁻⁹
Answer : The solubility of CaCO₃ is, 
Explanation :
As we know that CaCO₃ dissociates to give
ion and
ion.
The solubility equilibrium reaction will be:

The expression for solubility constant for this reaction will be,
![K_{sp}=[Ca^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
Let solubility of CaCO₃ be, 's'




Therefore, the solubility of CaCO₃ is, 
Answer:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Explanation:
Hello,
In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

The suitable equilibrium constant turns out:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Or in terms of the initial equilibrium constant:

Since the second reaction is a doubled version of the first one.
Best regards.