Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
Answer: Helianthus L. Sunflower
Explanation:
Answer: the coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.
Explanation:
Original volume of mercury = 1000 cm3.
The final volume of mercury considering its volume expansion quotient = 1000 + 1000*(1.8*10^-4 *52) = 1000 + 9.36 = 1009.36 cm^3
Considering the glass as a non expanding substance, the complete excess volume of 9.36 cm3 of mercury should have overflown the container, but due to the expansion of glass, the capacity of mercury containment increases and so a lesser amount of mercury flows out.
The amount of mercury that actually flowed out = 8.50 cm3.
So, the expansion of the glass container = 9.36-8.50 = 0.86 cm3.
Using the formula for coefficient of expansion,
coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.
Answer:
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Answer:
Option B. A
Explanation:
From the question given above, the following data were obtained:
C(s) + 2H₂ (g) —> CH₄ (g). ΔH = –74.9 kJ
From the reaction above, we can see that the enthalpy change (ΔH) is negative (i.e –74.9 KJ) which implies that the heat content of the reactants is greater than the heat content of the products. Thus, the reaction is exothermic reaction.
For an exothermic reaction, the energy profile diagram is drawn in such a way that the heat content of reactants is higher than the heat content of products because the enthalpy change
(ΔH) is always negative.
Thus, diagram A (i.e option B) gives the correct answer to the question.