Answer;
C7H14O2
Solution;
Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)
Mass of carbon = 12/44 × 2.726 g
= 0.743455 g
Mass of Hydrogen = 2/18 × 1.116 g
= 0.124 g
Mass of oxygen = 1.152 - (0.7435 + 0.124)
= 0.2845 g
Moles of carbon ; 0.7435/12 = 0.06196 moles
Moles of hydrogen; 0.124/1 = 0.124 moles
Moles of oxygen; 0.2845/16 = 0.01778 moles
Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778
= 3.5 : 7.0 : 1
To make them whole numbers ; we multiply the ratios by 2 to get;
(3.5 : 7.0 : 1 )2 = 7 : 14 : 2
Thus, the empirical formula of Isobutyl propionate is C7H14O2
Frequency.
The equation to find the velocity of a wave length is:
v=fλ
V stands for velocity
F stands for frequency
λ stands for wavelength
J. J. Thomson is the corect awncer
Density = 1.01 g/cm^3 or 1.01 kg/dm^3 or 1010 kg/m^3
Density = mass/volume = 1010 g/1000 cm^3 = 1.01 g/cm^3 = 1.01 kg/dm^3
= 1010 kg/m^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
