Answer:
ΔH= 3KJ
Explanation:
The total heat absorbed is the total energy in the process, and that is in form of entalpy.
ΔH = q + ΔHvap, where q is the heat necessary for elevate the temperature of dietil ether. Suppose the initial temperature is room temperature (25ºC=298 K), then
q= 10g x2.261 J/gK x(310 K - 298K)= 271.32 J= 0.3 kJ
Then
ΔHvap = 10g C4H10O x (1 mol C4H10O/74.12 g C4H10O) x( 15.7 KJ/ 1 mol C4H10O) = 2.12 KJ
ΔH= 2.5KJ ≈ 3KJ
Answer: 9.18 Litres
Standard Temperature and Pressure (STP). Think of this as the perfect environment where the Temp. is 0°C or 273 Kelvin and Pressure is always 1 atm. This is only true in STP.
This question uses the Ideal Gas Equation:
PV=nRT
P= 1 atm
V = ??
T = 273 K (always convert to Kelvin unless told otherwise)
n = 0.410 mol
R = 0.0821 L.atm/mol.K
What R constant to use depends on the units of the other values. (look at the attachments) The units cancel out and only Litres is left. You simply multiply the values.
Answer:
left to right across a period when it decreases and when it increases top to bottom in a group,
hope i helped
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.
