By sharing electeons with each other
if they lose or gain electrons then they only form ions
they cannot lose neutrons as they are locked inside the nucleus
Kindly checkout the attached pic. It contains all your required questions.
Answer:
You can see that the line is going up and is curved in a positive direction.
Explanation:
When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration.
Answer:
Mass = 279.23 g
Explanation:
Given data:
Number of moles of Fe₂O₃ = 3 mol
Number of moles of Al = 5 mol
Maximum amount of iron produced by reaction = ?
Solution:
Chemical equation:
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
Now we will compare the moles of iron with Al and iron oxide.
Fe₂O₃ : Fe
1 : 2
3 : 2×3 = 6 mol
Al : Fe
2 : 2
5 : 5 mol
The number of moles of iron produced by Al are less so Al is limiting reacting and it will limit the amount of iron so maximum number of iron produced are 5 moles.
Mass of iron:
Mass = number of moles × molar mass
Mass = 5 mol × 55.845 g/mol
Mass = 279.23 g
Answer:
a) The equilibrium will shift in the right direction.
b) The new equilibrium concentrations after reestablishment of the equilibrium :
![[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M](https://tex.z-dn.net/?f=%5BSbCl_5%5D%3D%280.370-x%29%20M%3D%280.370-0.0233%29%20M%3D0.3467%20M)
![[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BSbCl_3%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
![[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
Explanation:

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
On increase in amount of reactant

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of
is increasing .So, the equilibrium will shift in the right direction.
b)

Concentration of
= 0.195 M
Concentration of
= 
Concentration of
= 
On adding more
to 0.370 M at equilibrium :

Initially
0.370 M
At equilibrium:
(0.370-x)M
The equilibrium constant of the reaction = 

The equilibrium expression is given as:
![K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSbCl_3%5D%5BCl_2%5D%7D%7B%5BSbCl_5%5D%7D)

On solving for x:
x = 0.0233 M
The new equilibrium concentrations after reestablishment of the equilibrium :
![[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M](https://tex.z-dn.net/?f=%5BSbCl_5%5D%3D%280.370-x%29%20M%3D%280.370-0.0233%29%20M%3D0.3467%20M)
![[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BSbCl_3%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
![[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)