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 :
Answer:
no no no who are these some look good but are black what is this
Answer is B- He can arrange the beads into various formations using different colors and sizes multiple times
Explanation: I hope the helped!
H₂CO₃ ⇔ HCO₃⁻ + H⁺
I 0.160 0 0
C -x +x +x
E 0.160-x +x +x
Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]
4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)
x² = 6.88 x 10⁻⁸
x = 2.62 x 10⁻⁴
HCO₃⁻ ⇔ CO₃⁻² + H⁺
I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴
C -x +x +x
E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x
Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]
5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)
x = 5.6 x 10⁻¹¹
Thus,
[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M
[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M
[CO₃⁻²] = 5.6 x 10⁻¹¹ M
[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M
[OH⁻] = 3.8 x 10⁻¹¹
How will the candle contributes to the pressure and temperature of gases inside the glass?
The candle contributes to the pressure and temperature of the gases inside the glass because of the increasing temperature the lit of the candle disposes into the gas molecules inside and as these gas molecules rise in temperature they become unsettling and since heat makes gas molecules robust the area becomes smaller for them to be ecstatic and as the dynamics is explained p=f/a the pressure increases.
