Answer:
Decreasing the concentration of N2O3
Explanation:
This is because the products on the right of the reaction occupy more space. One (1) mole of NO and another mole of NO2 will occupy more space than the one (1) mole of N2O3. Therefore decreasing the concentration of N2O3 will shift the reaction to the right because the products will have more space to occupy – hence favoring equilibrium.
Answer:
0. 000115
Explanation:
A percentage is defined as a ratio with a basis of 100 as total substance. Convert a percentage to decimal implies to divide the percentage in 100 because decimal form has as basis 1.
For the isotopic forms:
1H: 99.98% → As percent.
99.98% / 100 = 0.9998 → As decimal form.
2H: 0.0115% → As percent.
0.0115% / 100 = <em>0. 000115</em>→ As decimal form.
Answer:
hydrogen builds many acids but not all
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
