Answer:
ethyl ethanoate and water
Explanation:
At the point when one fluid doesn't blend in with another yet glides on top of it, an isolating pipe can be utilized to isolate the two fluids. Oil glides on water. This combination can be isolated utilizing an isolating channel as demonstrated on the following page.
Ethyl liquor and water are two miscible fluids.
Refining is a cycle that can be utilized to isolate an unadulterated fluid from a combination of fluids. An isolating channel can be utilized to isolate the parts of the combination of immiscible fluids.
The answer is ethyl ethanoate and water. Hope this helps you!
Answer:
C
Explanation:
CH₄ is the formula for methane
Answer:
The new volume will be 367mL
Explanation:
Using PV = nRT
V1 = 259mL = 0.000259L
n1 = 0.552moles
At constant temperature and pressure, the value is
P * 0.000259 = 0.552 * RT ------equation 1
= 0.552 / 0.000259
= 2131.274
V2 = ?
n2 = 0.552 + 0.232
n2 = 0.784mole
Using ideal gas equation,
PV = nRT
P * V2 = 0.784 * RT ---------- equation 2
Combining equations 1 and 2 we have;
V2 = 0.784 / 2131.274
V2 = 0.000367L
V2 = 367mL
Explanation:
Reaction equation for this reaction is as follows.
It is given that 
= 0.0118.
According to the ICE table,
Initial: 0.86 0.86 0 0
Change: -x -x +x +x
Equilibrium: 0.86 - x 0.86 - x x x
Hence, value of will be calculated as follows.
0.0118 = 
x = 0.084 atm
Thus, we can conclude that 
is 0.084 atm.
