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Answer: Fluorine and chlorine are gases. Bromine is one of only two liquid elements, and iodine is a solid.
Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol
Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.
W are given a chemical reaction:



To calculate the enthalpy change, we use the formula:

This is the amount of energy released when 0.1326 grams of sample was burned.
So, energy released when 1 gram of sample was burned is = 
Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

Answer: It will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams
Explanation:
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Half life is represented by 

= rate constant
Given : Strontium-90 decreases in mass by one-half every 29 years , that is half life of Strontium-90 is 29 years.
As half life is independent of initial concentration, it will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams as the amount gets half.
Answer: Heat of vaporization is 41094 Joules
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:

where,
= initial pressure at 429 K = 760 torr
= final pressure at 415 K = 515 torr
= enthalpy of vaporisation = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 429 K
= final temperature = 515 K
Now put all the given values in this formula, we get
![\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B515%7D%7B760%7D%3D%5Cfrac%7B%5CDelta%20H%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B429K%7D-%5Cfrac%7B1%7D%7B415K%7D%5D)

Thus the heat of vaporization is 41094 Joules