Answer:
The simplified expression for the fraction is 
Explanation:
From the given information:
O3* → O3 (1) fluorescence
O + O2 (2) decomposition
O3* + M → O3 + M (3) deactivation
The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)
The rate of decomposition is = k₂ × cO
The rate of deactivation = k₃ × cO × cM
where cM is the concentration of the inert molecule
The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:
since cM is the concentration of the inert molecule
The leathery pad helps the camel because it prevents the camels feet from sinking into the sand.
Answer:
see explaination
Explanation:
Molecular equation;
2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)
Total ionic equation; . Includes all ions ;
6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)
Net ionic equation; remove common ions from total ionic;
2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)
The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
For more information about the partial pressure, refer to the link:
brainly.com/question/14623719
