439.3 g CO2
Explanation:
First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:
3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)
= 9.981 mol CO2
Use the molar mass of CO2 to determine the # of grams of CO2:
9.981 mol CO2 x (44.01 g CO2/1 mol CO2)
= 439.3 g CO2
Answer:49.3
Explanation:4.1j/g c * 25g * (t2-45c)=455j
T2-45c = 455j/4.1j/g c * 25g
455/104.6
45+4.3= 49.3 celsius
Answer:
The temperature of the Nitrogen after throttling is 
Explanation:
From the question we are told that
The temperature is 
The pressure is 
The pressure after being 
Generally from the first law of thermodynamics we have that

Here
is the change internal energy which is mathematically represented as

Here
is the specific heat of the gas at constant pressure
is the change kinetic energy which is negligible
Q is the thermal energy which is Zero for an adiabatic process
W is the work done and the value is zero given that the gas was throttled adiabatically
So

=> 
=> 
=> 
Answer:
Explanation:
CH₄ + H₂O(g) ⇒ CO(g)+3H2(g)
Equilibrium constant
K₁ = [CO][H₂]³ / [CH₄][H₂O]
CO(g)+H₂O(g) ⇒ CO₂(g) + H₂(g)
Equilibrium constant
K₂ = [CO₂][H₂] / [CO][H₂O]
CH₄(g)+2H₂O(g) ⇒ CO₂(g)+4H₂(g)
Equilibrium constant
K = [CO₂][H₂]⁴ /[CH₄][H₂O]²
= [CO][H₂]³ / [CH₄][H₂O] X [CO₂][H₂] / [CO][H₂O]
K₁ x K₂
K = K₁ x K₂