75.7 grams.
Convert moles to grams using this:
Moles • grams/1 mole = grams
(Grams is over 1 mole)
Hydrogen peroxide is H2O2, while water is H2O and oxygen (a diatomic gas) is O2. The (unbalanced) reaction is:
H2O2 --> H2O + O2
Notice that the H2O2 has 2 H atoms, and so does H2. This means that both must have the same coefficients, and we can adjust the coefficient of O2. Since H2O2 has 2 O atoms, and H2O has 1, we multiply O2 by 1/2:
H2O2 --> H2O + (1/2)O2
This has an equivalent number of H and O atoms on either side, but we want the coefficients to be whole numbers, so we multiply everything by 2:
2H2O2 --> 2H2O + O2
I’m not totally sure but I think the answer is B
I think it is eutrophication
<u>Answer:</u> The activation energy for the reaction is 40.143 kJ/mol
<u>Explanation:</u>
To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:
![\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B317K%7D%7D%7BK_%7B278K%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 317 K = 
= equilibrium constant at 278 K = 
= Activation energy = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = 278 K
= final temperature = 317 K
Putting values in above equation, we get:
![\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B3.050%5Ctimes%2010%5E8%7D%7B3.600%5Ctimes%2010%5E%7B7%7D%7D%29%3D%5Cfrac%7BE_a%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B278%7D-%5Cfrac%7B1%7D%7B317%7D%5D%5C%5C%5C%5CE_a%3D40143.3J%2Fmol%3D40.143kJ%2Fmol)
Hence, the activation energy for the reaction is 40.143 kJ/mol