Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
I believe the density p1 is greater than the density p2 .
Since the liquid are at equilibrium in the the open U-tube, the pressure at which the liquids meet should be the same. That is at the position where they are in contact, the pressure that liquid 1 exerts at that point is the same as the pressure exerted by liquid 2 at the point.
Answer: You can probably go onto <em><u>You Tube and find that answer!</u></em> Just be <u><em>detailed in your search!</em></u> Hope this helps!
Explanation:
Just Simple Physics :P
