Answer:
The ratio of f at the higher temperature to f at the lower temperature is 5.356
Explanation:
Given;
activation energy, Ea = 185 kJ/mol = 185,000 J/mol
final temperature, T₂ = 525 K
initial temperature, T₁ = 505 k
Apply Arrhenius equation;
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
Where;
is the ratio of f at the higher temperature to f at the lower temperature
R is gas constant = 8.314 J/mole.K
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7B185%2C000%7D%7B2.303%20%5Ctimes%208.314%7D%20%5B%5Cfrac%7B1%7D%7B505%7D%20-%5Cfrac%7B1%7D%7B525%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%200.7289%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%2010%5E%7B0.7289%7D%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%205.356)
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356
Answer: 13.31 moles.
Explanation: So take 452 grams of Argon and multiply by the molar mass of Argon. Your units will cancel out, leaving you with moles of Argon.
10 decigrams are equivalent to 1 gram, so 2,100 grams are equivalent to 2100 x 10. This makes 21,000.
Answer:
Explanation:
the given equation is not balance
the balance equation can be written as
4P+5O2...........2P2O5
