In which type of radioactive decay does the nucleus become more stable without changing its identity?(1 point)
2 answers:
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Answer:
Total pressure at equilibrium is 0.2798atm.
Explanation:
For the reaction:
H₂S(g) ⇄ H₂(g) + S(g)
Kp is defined as:
If initial pressure of H₂S is 0.150 atm, equilibrium pressures are:
H₂S(g): 0.150atm - x
H₂(g): x
S(g): x
Replacing in Kp:
X² = 0.1251 - 0.834X
X² + 0.834X - 0.1251 = 0
Solving for X:
X = -0.964 → False solution: There is no negative pressures
X = 0.1298
Thus, pressures are:
H₂S(g): 0.150atm - 0.1298atm = <em>0.0202atm</em>
H₂(g): <em>0.1298atm</em>
S(g): <em>0.1298atm</em>
Thus, total pressure in the container at equilibrium is:
0.0202atm + 0.1298atm + 0.1298atm = <em>0.2798atm</em>