Answer:
5.0×1022 sa taas po yang 22
Answer:
At sea level, the atmospheric pressure would be a little over 100 kPa (one atmosphere or 760 mm Hg). If we climb to the top of Mount Everest (the highest mountain in the world at 29,029 feet or 8848 meters), the atmospheric pressure will drop to slightly over 30 kPa (about 0.30 atmospheres or 228 mm Hg).
Answer:
A. ![r=k[NO]^2[H_2]](https://tex.z-dn.net/?f=r%3Dk%5BNO%5D%5E2%5BH_2%5D)
B. ![k=0.121\frac{L^2}{mol^2*s}](https://tex.z-dn.net/?f=k%3D0.121%5Cfrac%7BL%5E2%7D%7Bmol%5E2%2As%7D)
C. Second-order with respect to NO and first-order with respect to H₂
Explanation:
Hello,
In this case, for the reaction:
![H_2O(g) + 2NO(g) \rightarrow N_2O + H_2O(g)](https://tex.z-dn.net/?f=H_2O%28g%29%20%2B%202NO%28g%29%20%5Crightarrow%20N_2O%20%2B%20H_2O%28g%29)
The rate law is determined by writing the following hypothetical rate laws:
![3.822x10^{-3}=k[0.3]^m[0.35]^n\\\\1.529x10^{-2}=k[0.6]^m[0.35]^n\\\\3.058x10^{-2}=k[0.6]^m[0.7]^n](https://tex.z-dn.net/?f=3.822x10%5E%7B-3%7D%3Dk%5B0.3%5D%5Em%5B0.35%5D%5En%5C%5C%5C%5C1.529x10%5E%7B-2%7D%3Dk%5B0.6%5D%5Em%5B0.35%5D%5En%5C%5C%5C%5C3.058x10%5E%7B-2%7D%3Dk%5B0.6%5D%5Em%5B0.7%5D%5En)
Whereas we can compute m as follows:
![\frac{3.822x10^{-3}}{1.529x10^{-2}} =\frac{[0.3]^m[0.35]^n}{[0.6]^m[0.35]^n} \\\\0.25=(0.5)^m\\\\m=\frac{log(0.25)}{log(0.5)} \\\\m=2](https://tex.z-dn.net/?f=%5Cfrac%7B3.822x10%5E%7B-3%7D%7D%7B1.529x10%5E%7B-2%7D%7D%20%3D%5Cfrac%7B%5B0.3%5D%5Em%5B0.35%5D%5En%7D%7B%5B0.6%5D%5Em%5B0.35%5D%5En%7D%20%5C%5C%5C%5C0.25%3D%280.5%29%5Em%5C%5C%5C%5Cm%3D%5Cfrac%7Blog%280.25%29%7D%7Blog%280.5%29%7D%20%5C%5C%5C%5Cm%3D2)
Therefore, the reaction is second-order with respect to NO. Thus, for hydrogen, we find n:
![\frac{1.529x10^{-2}}{3.058x10^{-2}} =\frac{[0.6]^2[0.35]^n}{[0.6]^2[0.7]^n} \\\\0.5=(0.5)^n\\\\n=\frac{log(0.5)}{log(0.5)}\\ \\n=1](https://tex.z-dn.net/?f=%5Cfrac%7B1.529x10%5E%7B-2%7D%7D%7B3.058x10%5E%7B-2%7D%7D%20%3D%5Cfrac%7B%5B0.6%5D%5E2%5B0.35%5D%5En%7D%7B%5B0.6%5D%5E2%5B0.7%5D%5En%7D%20%5C%5C%5C%5C0.5%3D%280.5%29%5En%5C%5C%5C%5Cn%3D%5Cfrac%7Blog%280.5%29%7D%7Blog%280.5%29%7D%5C%5C%20%5C%5Cn%3D1)
A) Therefore, the reaction is first-order with respect to H₂. In such a way, we conclude that that the rate law is:
![r=k[NO]^2[H_2]](https://tex.z-dn.net/?f=r%3Dk%5BNO%5D%5E2%5BH_2%5D)
B) Rate constant is computed from one kinetic data:
![k=\frac{1.529x10^{-2}\frac{mol}{L*s} }{(0.6\frac{mol}{L} )^2(0.35\frac{mol}{L})}\\\\k=0.121\frac{L^2}{mol^2*s}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1.529x10%5E%7B-2%7D%5Cfrac%7Bmol%7D%7BL%2As%7D%20%7D%7B%280.6%5Cfrac%7Bmol%7D%7BL%7D%20%29%5E2%280.35%5Cfrac%7Bmol%7D%7BL%7D%29%7D%5C%5C%5C%5Ck%3D0.121%5Cfrac%7BL%5E2%7D%7Bmol%5E2%2As%7D)
C. As mentioned before, reaction is second-order with respect to NO and first-order with respect to H₂.
Best regards.
Answer:
An infographic is a collection of imagery, charts, and minimal text that gives an easy-to-understand overview of a topic. As in the example below, infographics use striking, engaging visuals to communicate information quickly and clearly. Infographics are a valuable tool for visual communication.
Answer:
190 g
Explanation:
Knowing that:
- O₂ gas in air: 0.28 g O₂ / L of air
- Air inhaled at rest: 19.3 fl.oz. air / breath
- Human takes 20 breaths / min
The grams of oxygen gas (O₂) that a human at rest inhales in 1 hour (60 minutes) will be calculated as:
grams of O₂ = breaths in 1 hour * air inhaled per breath * amount of O₂ in air
![grams O_{2} = 60 min* 20 \frac{breath}{min} * 19.3 \frac{fl.oz. air}{breath} * \frac{1 L}{33.814 fl.oz.} * 0.28 \frac{g O_{2} }{L}](https://tex.z-dn.net/?f=grams%20O_%7B2%7D%20%3D%2060%20min%2A%2020%20%5Cfrac%7Bbreath%7D%7Bmin%7D%20%2A%2019.3%20%5Cfrac%7Bfl.oz.%20air%7D%7Bbreath%7D%20%2A%20%5Cfrac%7B1%20L%7D%7B33.814%20fl.oz.%7D%20%2A%200.28%20%5Cfrac%7Bg%20O_%7B2%7D%20%7D%7BL%7D)
grams of O₂ = 191.78 g
190 g of O₂ are inhaled in 1 hour (written in correct number of significant figures)
*In the calculation the amount of air inhaled per breath was converted from fl.oz. to Liters knowing that <em>1 L = 33.814 fl.oz.</em>