Answer:
The answer is b, c, d, e
Explanation:
b. 2 N2O5 → 4 NO2 + O2
r = k [N2O5]^2 --> Second-order regarding global reaction
c. 2 HI → H2 + I2
r = k [HI]^2 --> Second-order regarding global reaction
d. 2 N2O → 2 N2 + O2
r = k [N2O]^2 --> Second-order regarding global reaction
e. 2 NO2 → 2 NO + O2
r = k [NO2]^2 --> Second-order regarding global reaction
C! objects that carry instruments into the stratosphere to measure atmospheric conditions
Answer:
False.
Explanation:
<u>The given statement asserts a false claim because the equator is the region that receives maximum sunlight</u>. The equator is placed right below the sun and thus, it tends to receive the maximum radiation across the year. While the poles are the coldest regions of the Earth because due to Earth's titled axis, they receive very few sun rays for a certain time of the year. Thus, <u>if we move away from the equator, we are likely to receive less radiation from the sun as the sun keeps getting farther while moving away from the equator</u>.
Nectar contains a high concentration of sucrose (sugar).
When hummingbirds ingest the nectar, their cells use oxygen to convert it into carbon dioxide and water.
The oxidation of sucrose releases a large amount of energy, which the birds can use for flapping their wings, etc.
Answer : The temperature will be, 392.462 K
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
=
= rate constant at 
= 
= activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 293 K
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B3K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B66410J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B293K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
Therefore, the temperature will be, 392.462 K
