<span>So to make it clear let's break the equation down species by species and assess the number of each species on bothe sides of the equation:
2C</span>₈H₈ + 25O₂ → 8CO₂ + 18H₂<span>O
LHS: C - 16 RHS: C - 8
H - 16 H - 36
O - 50 O - 34
Thus based on that it is evident that the equation is not quite balanced. This therefore means a "</span><span>No, because the number of carbon, hydrogen & oxygen atoms on both sides of the equation are not equal."
</span>The actual balance equation would be C₈H₈ + 10O₂ → 8CO₂ + 4H₂O
I think you meant K as in kilograms?
If so, the answer should be 0.0052.
Mass number = protons + neutrons. 12 + 11 = 23.
pH=2.7
<h3>Further explanation</h3>
Acetic acid = weak acid
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Ka = acid ionization constant
M = molarity
Ka for Acetic acid(CH₃COOH) : 1.8 x 10⁻⁵
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.222}\\\\=0.001998=1.998\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.222%7D%5C%5C%5C%5C%3D0.001998%3D1.998%5Ctimes%2010%5E%7B-3%7D)
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.
