If 0.5 L of solution contains 4 mol
then let 1 L of solution contain x mol
⇒ (0.5 L) x = (4 mol) (1 L)
x = (4 mol · L) ÷ (0.5 L)
x = 8 mol
Thus the molarity of the Sodium Chloride solution is 8 mol / L OR 8 mol/dm³.
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
Answer:
Physical properties are controlled by the mineral's chemical composition and atomic structure, this combination is distinct in each mineral.
Explanation:
Hello,
Since the physical properties depend on the mineral's chemical composition and the atomic structure (molecular arrangement), we can distinguish different minerals by measuring out hardness, cleavage and other properties in order to analyze their behavior under several conditions and subsequently use them in a particular application.
Best regards.
=
= 0.134 mol
mol of oxygen:
ratio of
= 2 : 15
= 1 : 7.5
: . mol of
= 0.134mol * 7.5
= 1.01 mol
Mass of Oxygen = mol * Mr
= 1.01 mol * (16*2) g/mol
= 32.22g
Note: Mr is molar mass
