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!
Explanation:
but.......you havent attached a picture?
<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
Answer:
14 mol O₂
Explanation:
The reaction between CO and O₂ is the following:
CO + O₂ → CO₂
We balance the equation with a coefficient 2 in CO and CO₂ to obtain the same number of O atoms:
2CO + O₂ → 2CO₂
As we can see from the balanced equation, 1 mol of O₂ is required to react with 2 moles of CO. Thus, the conversion factor is 1 mol of O₂/2 mol CO. We multiply the moles of CO by the conversion factor to calculate the moles of O₂ that are required:
28 mol CO x 1 mol of O₂/2 mol CO = 14 mol O₂
Answer:
If u use math why It will give u the answer like 3x3=9
Explanation:
