Answer:
In thermodynamics, the Joule–Thomson effect describes the temperature change of a real gas or liquid when it is forced through a valve or porous plug while keeping it insulated so that no heat is exchanged with the environment. This procedure is called a throttling process or Joule–Thomson process
Answer : At constant pressure work is done by the system on the surroundings.
Explanation :
Work done : Any quantity that flows across the boundary of a system during a change in its state and it completely convertible into the lifting of a weight in the surroundings.
Formula for work done is:

Sign convention :
- When volume expand then system work that means work done by the system.
w = (-ve)
- When volume compress then surrounding work that means work done on the system.
w = (+ve)
The given reaction is:

This is a evaporation process in which phase changes from liquid state to gaseous state at constant temperature.
At constant pressure, work depends only on volume.
In evaporation process, the volume expand that means work is done by the system on the surroundings.
Sign convention is, w = (-ve)
Thus, at constant pressure work is done by the system on the surroundings.
Answer:
Explanation:
C₂H₂ + 2H₂ = C₂H₆
1 mole 2 mole 1 mole
Feed of reactant is 1.6 mole H₂ / mole C₂H₂
or 1.6 mole of H₂ for 1 mole of C₂H₂
required ratio as per chemical reaction written above
2 mole of H₂ for 1 mole of C₂H₂
So H₂ is in short supply . Hence it is limiting reagent .
1.6 mole of H₂ will react with half of 1.6 mole or .8 mole of C₂H₂ to form .8 mole of C₂H₆
a )Calculate the stoichiometric reactant ratio = mole H₂ reacted/mole C₂H₂ reacted
= 1.6 / .8 = 2 .
b )
yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .
Mg + 1/2 O2 → MgO
1 mol = 24 g of Mg
X mol = 12 g of Mg
x = 0.5 moles of Mg
Mg :MgO = 1:1 (coefficient from equations using mole ratio)
So
0.5 moles of MgO
1 mol MgO = (24+16) g = 40 g
0.5 moles of MgO = 0.5 × 40
= 20 g of MgO produced
Answer:
Look at the picture.
Explanation:
On stage one binding of a substrate occurs (and also the geometry of active site may change) and water comes to the site. On stage two the hydrolisis takes place and on stage 3 products deabsorb from the enzyme.