4. According to the explanation given in one previous session, in this case we have the following reaction:
Zn + NO3- -> Zn(OH)42- + NH3
Our important informations to have in mind always first:
Single elements = 0 of oxidation number
Zn = 0
Oxygen = 2-
NO3- = overall charge must be -1, and we have 3 oxygens, so a -6 charge, therefore N must give a charge of +5
N (reactant) = 5+
Zn(OH)42- = overall charge -2, the charge for OH is always -1, we have 4 OH, therefore -4 of charge, to give -2 of overall, Zn must have a charge of +2
Zn (product) = 2+
OH = 1-
NH3 = 0
Now for the balancing of the reaction:
4 Zn + NO3- + 7 OH- + 6 H2O -> NH3 + 4 Zn(OH)42-
Answer:To be honest i dont know
Explanation:
e
Answer:
Option c. 2Kl + Cl2 —> 2KCl + l2
Explanation:
A balanced equation must have the same number of atoms on the reactants side and the products side. This do the fact of the postulation by the mass conservation principle that says: Atoms can neither be broken nor destroyed but can be arranged. The atoms on the left side for each element are equal. Thus, C presents a balanced equation.
Answer:
Of the following equilibria, only one will shift to the right in response to a decrease in volume.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
Explanation:
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.
Decrease the volume
If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.
On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.
On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.