Answer:
0.66atm
Explanation:
Based on the reaction:
NH₄HS(s) ↔ NH₃(g) + H₂S(g)
The equilibrium constant, Kp, is defined as:
Kp = 0.11 = 
As moles of gas produced for NH₃(g) and H₂S(g) are the same, it is possible to write:


That means pressure of NH₃(g) is 0.33atm and H₂S(g) is, also, 0.33atm. Thus, total pressure is:
0.33atm×2 = <em>0.66atm</em>
Answer:
2800 g de ácido nítrico
Explanation:
La ecuación por la oxidación de amoniaco es:
4NH₃ + 7O₂ → 4H₂O + 2HNO₂ + 2HNO₃
Si pensamos que el oxígeno es el reactivo limitante, trabajamos con el amoniaco. Convertimos su masa a moles:
1.36 kg = 1360 g
1360 g . 1mol /17g = 80 moles
Si 4 moles de amoniaco pueden producir 2 moles de acido nítrico
80 moles producirán, (80 . 2)/4 = 40 moles.
Convertimos los moles a gramos:
40 mol . 63g /mol = 2520 g
Si le aplicamos la pureza
2520 g . 100/90 = 2800 g
Explanation:
The required concentration of
M1 =0.222 M.
The required volume of
is V1 =225 mL.
The standard solution of
is M2 =16 M.
The volume of standard solution required can be calculated as shown below:
Since the number of moles of solute does not change on dilution.
The number of moles 


Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.
The answer to this question would be: 2 mol
To answer this question, you need to know the molecular weight of Potassium. Molecular weight determines how much the weight of 1 mol of a molecule has.
Potassium or Kalium molecular weight is 39.1 gram/mol. Then, 78.20gram of potassium should be: 78.20g/ (39.1g/mol)= 2 mol