Answer: K = 284
Explanation:
Gibbs free energy (DG) = RT(InK)
Where :
R = Gas constant = 8.314J/K/mol = 0.008314KJmol/K
T = temperature in Kelvin = 298K
K = dissociation rate
DG = 14
DG = RT(InK)
14 = 0.008314 × 298 × (InK)
14 = 2.477572 × InK
Ink = (14 ÷ 2.477542)
InK = 5.6507619
K = 284
Answer: The ground-state electronic configuration of
1) ![Ru^{2+}=[Kr]4d^6](https://tex.z-dn.net/?f=Ru%5E%7B2%2B%7D%3D%5BKr%5D4d%5E6)
2) ![W^{3+]=4f^{14}5d^3](https://tex.z-dn.net/?f=W%5E%7B3%2B%5D%3D4f%5E%7B14%7D5d%5E3)
Explanation: The electronic configuration of elements is defined as the distribution of electrons around the nucleus of that element. It depends on the atomic number of the element.
Atomic number of the element = Number of electrons.
Atomic number = 44
Number of electrons = 44
Electronic configuration of Ru-element = ![[Kr]4d^75s^1](https://tex.z-dn.net/?f=%5BKr%5D4d%5E75s%5E1)
To form
, 2 electrons are released from the neutral Ru-element.
So, the electronic configuration of ![Ru^{2+}=[Kr]4d^6](https://tex.z-dn.net/?f=Ru%5E%7B2%2B%7D%3D%5BKr%5D4d%5E6)
Atomic number = 74
Number of electrons = 74
Electronic configuration of W-element = ![[Xe]4f^{14}5d^46s^2](https://tex.z-dn.net/?f=%5BXe%5D4f%5E%7B14%7D5d%5E46s%5E2)
To form
, 3 electrons are released from the neutral W-element.
So, the electronic configuration of ![W^{3+}=[Xe]4f^{14}5d^3](https://tex.z-dn.net/?f=W%5E%7B3%2B%7D%3D%5BXe%5D4f%5E%7B14%7D5d%5E3)
Answer:
The equilibrium concentration of
.
The equilibrium concentration of
.
The equilibrium concentration of
.
Explanation:
Answer:
The equilibrium concentration of HCl is 0.01707 M.
Explanation:
Equilibrium constant of the reaction = 
Moles of 
Concentration of ![[PCl_3]=\frac{0.280 mol}{1.00 L}=0.280 M](https://tex.z-dn.net/?f=%5BPCl_3%5D%3D%5Cfrac%7B0.280%20mol%7D%7B1.00%20L%7D%3D0.280%20M)
Moles of 
Concentration of ![[Cl_2]=\frac{0.280 mol}{1.00 L}=0.280M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D%5Cfrac%7B0.280%20mol%7D%7B1.00%20L%7D%3D0.280M)

Initial: 0.280 0.280 0
At eq'm: (0.280-x) (0.280-x) x
We are given:
![[PCl_3]_{eq}=(0.280-x)](https://tex.z-dn.net/?f=%5BPCl_3%5D_%7Beq%7D%3D%280.280-x%29)
![[Cl_2]_{eq}=(0.280-x)](https://tex.z-dn.net/?f=%5BCl_2%5D_%7Beq%7D%3D%280.280-x%29)
![[PCl_5]_{eq}=x](https://tex.z-dn.net/?f=%5BPCl_5%5D_%7Beq%7D%3Dx)
Calculating for 'x'. we get:
The expression of
for above reaction follows:
![K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BPCl_3%5D%5BCl_2%5D%7D)
Putting values in above equation, we get:

On solving this quadratic equation we get:
x = 0.228, 0.344
0.228 M < 0.280 M< 0.344 M
x = 0.228 M
The equilibrium concentration of
.
The equilibrium concentration of
.
The equilibrium concentration of
.
Answer:
phenotype,phenotype,genotype,genotype,
genotype
Explanation:
phenotype is physical appearance and genotype is just like
yy Tt
<span>The effective nuclear charge of an atom = total electrons - inner electrons
For O, ENC = 8 - 2 = 6
For Li, ENC = 3 - 2 = 1
For C, ENC = 6 - 2 = 4
The electrons in O experience the greatest effective nuclear charge and that is why O is smaller than C (which is smaller than Li).</span>