We need to first add both of the solution volumes together 35+115=150. Now we can divide the volume of the ethanol by the total volume 35/150=.233. To double check we can multiply the total volume by the percentage of ethanol by volume we got as a solution 150x.233=35. So the percentage by volume of ethanol in the solution is .233x100=23.3%.
Answer : The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2
Explanation :
The relation between the equilibrium constant and standard Gibbs free energy is:
![\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Ctimes%20%5Cln%20Q%5C%5C%5C%5C%5CDelta%20G%5Eo%3D-RT%5Ctimes%20%5Cln%20%28%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%29)
where,
= standard Gibbs free energy = -14.1 kJ/mol
R = gas constant = 8.314 J/K.mol
T = temperature = 
Q = reaction quotient
![[A]_{inside}](https://tex.z-dn.net/?f=%5BA%5D_%7Binside%7D)
= concentration inside the cell
![[A]_{outside}](https://tex.z-dn.net/?f=%5BA%5D_%7Boutside%7D)
= concentration outside the cell
Now put all the given values in the above formula, we get:
![-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})](https://tex.z-dn.net/?f=-14.1%5Ctimes%2010%5E3J%2Fmol%20%3D-%288.314J%2FK.mol%29%5Ctimes%20%28298K%29%5Ctimes%20%5Cln%20%28%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%29)
![\frac{[A]_{inside}}{[A]_{outside}}=296.2](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_%7Binside%7D%7D%7B%5BA%5D_%7Boutside%7D%7D%3D296.2)
Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2
<span>Water molecules form a complex with metal ions (usually a 6-coordinate complex). And the high charge density on a metal ion draws electrons away from the water molecules, making the O-H bonds more polar than normal. This allows the dissociation of the protons, making solutions of most metal ions acidic</span>
Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(NH4)2CO3) = ?
For the purpose, here we will use the next equation:
c=n/V ⇒ n=cxV
n((NH4)2CO3) = 0.02M x 0.001L
n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution
Octane is the resistance of the burning of gasoline
