Answer : The value of equilibrium constant (Kc) is, 0.0154
Explanation :
The given chemical reaction is:

Initial conc.
0 0
At eqm.
x x
As we are given:
Concentration of
at equilibrium = 
That means,

The expression for equilibrium constant is:
![K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSO_2%5D%5BCl_2%5D%7D%7B%5BSO_2Cl_2%5D%7D)
Now put all the given values in this expression, we get:



Thus, the value of equilibrium constant (Kc) is, 0.0154
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
Cu + 2H2SO4 ⟶ CuSO4 + SO2 + 2H20
In left hand side of the equation.
Cu = 1 atom
H = 4 atoms
S = 2 atoms
O = 8 atoms
In right hand side of the equation.
Cu = 1 atom
S = 2 atoms
0 = 8 atoms
H = 4 atoms
• All the atoms are balanced in the left and right side of the equation and it satisfies the law of conservation of mass.
• Equation is balanced and correct.
*See the attachment .
Answer:
more rounded the grains are the more they have been moved around
Explanation:
Generally – the more rounded the grains are the more they have been moved around (i.e. the longer the length of time or distance they have moved). Angular grains cannot have travelled far
geolsoc.org.uk
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15