1. <span>EACH ORBITS HOLDS A FIXED NUMBER OF ELECTRONS </span>
<u>Answer:</u> The activation energy of the reaction is 124.6 kJ/mol
<u>Explanation:</u>
To calculate activation energy of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B79%5EoC%7D%7D%7BK_%7B26%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 79°C = 
= equilibrium constant at 26°C = 
= Activation energy of the reaction = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![26^oC=[26+273]K=299K](https://tex.z-dn.net/?f=26%5EoC%3D%5B26%2B273%5DK%3D299K)
= final temperature = ![79^oC=[79+273]K=352K](https://tex.z-dn.net/?f=79%5EoC%3D%5B79%2B273%5DK%3D352K)
Putting values in above equation, we get:
![\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B0.394%7D%7B2.08%5Ctimes%2010%5E%7B-4%7D%7D%29%3D%5Cfrac%7BE_a%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B299%7D-%5Cfrac%7B1%7D%7B352%7D%5D%5C%5C%5C%5CE_a%3D124595J%2Fmol%3D124.6kJ%2Fmol)
Hence, the activation energy of the reaction is 124.6 kJ/mol
A 3.0 L balloon is 2.5 times as big as a 1.2L balloon, and 3 times .060 equals .18, so if you multiply .18 by 2.5 you get .45 moles of air i’m pretty sure
Explanation:
1 mole = 6.02 x 10^23 atoms (Avogadro’s number)
Step 1) Determine how many grams of a substance are in the problem
Step 2) Find the amount of grams in 1 mole of the substance
3) Multiply step one by step two
