Explanation:
option A skeleton system is correct option
hope this helps you !
Answer: weigh is m = n × M = 2.87 mol × 58.44 g/mol
Explanation: mass = amount of substance × molar mass
M((NaCl) = 22.99 +35.45
Answer:
The one that is oxidized or the reducing agent
Explanation:
Oxidation results in loss of electrons or an increase in oxidation state.
Reduction results in gain of electrons or decrease in oxidation state.
The element that undergoes oxidation is said to oxidised, similarly the element that undergoes reduction is said to be reduced.
In a redox reaction, both oxidation and reduction takes place.
If a substrate undergoes oxidation or is oxidised, it is also responsible for reduction of the other species, as the total number of electrons should always be conserved.
The substance that undergoes oxidation, releases some electrons, these electrons are taken by the other substrate and it undergoes reduction.
Hence the substance that undergoes oxidation is called the reducing agent, as it is responsible for reduction of other substrate, when oxidizing itself.
Similar thought works for a substance that undergoes reduction or is reduced, works as an oxidizing agent.
Answer:
4. 60 neutrons.
Explanation:
The given isotopes;
¹⁰⁶₄₆Pd
In this isotope, we can deduce that the mass number is the superscript and the atomic number is the subscript;
Mass number = 106
Atomic number = 46
Mass number is the number of protons and neutrons in an atom;
Mass number = Protons + neutrons
Atomic number is the number of protons
So, Number of protons = 46
Number of neutrons = Mass number - Atomic number
= 106 - 46
= 60
Number of neutrons = 60
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 :
