None since CO3 does not exist.
Answer:
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Explanation:
For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.
For a hypothetical reaction:
xA + yB ⇄ zC
The equilibrium constant is :
![Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%7D%7B%5BC%5D%5E%7Bz%7D%20%7D)
The given reaction involves the decomposition of H2O into H2 and O2
![2H_{2}O\rightleftharpoons 2H_{2} + O_{2}](https://tex.z-dn.net/?f=2H_%7B2%7DO%5Crightleftharpoons%202H_%7B2%7D%20%2B%20O_%7B2%7D)
The equilibrium constant is expressed as :
![Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Since Keq = 5.31*10^-10
![5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}](https://tex.z-dn.net/?f=5.31%2A10%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B%5DH_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D%7B%5BH_%7B2%7DO%5D%5E%7B2%7D%7D)
Answer:
false
Explanation:
Atoms of the same element that differ in their numbers of neutrons are called isotopes. Many isotopes occur naturally. ... Different isotopes of an element generally have the same physical and chemical properties because they have the same numbers of protons and elec
Answer:
![\Delta _fH_{C_3H_8}=-102.7kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20_fH_%7BC_3H_8%7D%3D-102.7kJ%2Fmol)
Explanation:
Hello,
In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:
![\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}](https://tex.z-dn.net/?f=%5CDelta%20_cH%3D3%2A%5CDelta%20_fH_%7BCO_2%7D%2B4%2A%5CDelta%20_fH_%7BH_2O%7D-%5CDelta%20_fH_%7BC_3H_8%7D-5%2A%5CDelta%20_fH_%7BO_2%7D)
Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:
![\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20_fH_%7BC_3H_8%7D%3D3%2A%5CDelta%20_fH_%7BCO_2%7D%2B4%2A%5CDelta%20_fH_%7BH_2O%7D-5%2A%5CDelta%20_fH_%7BO_2%7D-%5CDelta%20_cH%5C%5C%5C%5C%5CDelta%20_fH_%7BC_3H_8%7D%3D3%2A%28-393.5%29%2B4%2A%28-241.8%29-5%2A0-%28-2045%29%5C%5C%5C%5C%5CDelta%20_fH_%7BC_3H_8%7D%3D-102.7kJ%2Fmol)
Best regards.
Answer:
are the best type of scientific investigation to demonstrate cause-and-effect relationships because they allow the investigator to actively manipulate variables and control conditions.