Answer:
Please refer to the attachment for answers.
Explanation:
Please refer to the attachment for explanation
Answer:
Group of highly-reactive chemical elements. The alkali metals are a group (column) in the periodic table consisting of the chemical elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr).
Answer:
The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases
Explanation:
Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in solution decreases significantly.
In this case, both BaSO4 and Na2SO4 both possess the SO4^2- anion. Hence SO4^2- anion is the common ion. Given the equilibrium;
BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq), addition of Na2SO4 will decrease the solubility of BaSO4 due to the presence of a common SO4^2- anion compared to pure water.
This implies that the equilibrium will shift to the left, (more undissoctiated BaSO4) hence decreasing the Ba^2+(aq) concentration.
The atomic radius increases down a column (group) and decreases along a row
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
