Answer:
4.75 is the equilibrium constant for the reaction.
Explanation:
Equilibrium concentration of reactants :
![[CO]=0.0590 M,[H_2O]=0.00600 M](https://tex.z-dn.net/?f=%5BCO%5D%3D0.0590%20M%2C%5BH_2O%5D%3D0.00600%20M)
Equilibrium concentration of products:
![[CO_2]=0.0410 M,[H_2]=0.0410 M](https://tex.z-dn.net/?f=%5BCO_2%5D%3D0.0410%20M%2C%5BH_2%5D%3D0.0410%20M)
The expression of an equilibrium constant is given by :
![K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)
4.75 is the equilibrium constant for the reaction.
Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
<em></em>
Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere = 
==> V = 
= 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>
The atomic structure of the atom contains 9 positively charged particles (protons) and 10 neutrally charged particles (neutrons) in the center of the atom in a clump called the nucleus. Those 9 negatively charged particles (electrons) are moving around outside of the nucleus.
There are 10 neutral charges, because the mass of 19 comes from the number of neutral charges plus the number of positive charges.
To calculate the number of neutral charges, subtract the positive charges from the mass (19 - 9), and you get the number of neutral charges (10).
%C= 12/12 + 2·16=0,273=27,3%.
