Answer:
A) = 4.7 × 10⁻⁴atm
Explanation:
Given that,
Kp = 1.5*10³ at 400°C
partial pressure pN2 = 0.10 atm
partial pressure pH2 = 0.15 atm
To determine:
Partial pressure pNH3 at equilibrium
The decomposition reaction is:-
2NH3(g) ↔N2(g) + 3H2(g)
Kp = [pH2]³[pN2]/[pNH3]²
pNH3 =√ [(pH2)³(pN2)/Kp]
pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm
![K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cfrac%7B%5BpH_2%5D%20%5E3%5BpN_2%5D%7D%7B%5BpNH_3%5D%5E2%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%28pH_2%29%5E3%28pN_2%29%7D%7BpNH_3%7D%20%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.15%29%5E3%280.10%29%7D%7B1.5%20%5Ctimes%2010%5E3%7D%20%7D%20%5C%5C%3D4.74%20%5Ctimes%2010%5E-%5E4atm)
= 4.7 × 10⁻⁴atm
C. Digesting a sandwich
Is your answer
Hope this helps
Answer:
2%
Explanation:
.98 is 98% of one and therefore they are missing 2%
Place a burning splint near the opening of a test tube. If a popping noise occurs, it's probably hydrogen. Place a glowing splint in the test tube, and if it reignites, it could be oxygen. Place a burning splint into a test tube, and if it goes out, it could be carbon dioxide.
The rate of a reaction would be one-fourth.
<h3>Further explanation</h3>
Given
Rate law-r₁ = k [NO]²[H2]
Required
The rate of a reaction
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
Can be formulated:
Reaction: aA ---> bB

or

The concentration of NO were halved, so the rate :
![\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1](https://tex.z-dn.net/?f=%5Ctt%20r_2%3Dk%5B%5Cdfrac%7B1%7D%7B2%7DNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dk.%5BNo%5D%5E2%5BH_2%5D%5C%5C%5C%5Cr_2%3D%5Cdfrac%7B1%7D%7B4%7Dr_1)