It is the bottom answer. (77.5)
If the current can’t move the tree, the tree stop the current
Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
M(Ag)=12.5 g
Nₐ=6.022*10²³ mol⁻¹
n(Ag)=m(Ag)/M(Ag)
N=n(Ag)*Nₐ
N=Nₐm(Ag)/M(Ag)
N=6.022*10²³mol⁻¹*12.5g/107.868g/mol=6.97*10²²
A. 6.97 × 10²² atoms
Answer:
Q = 143,921 J = 143.9 kJ.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

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