Answer:
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Explanation:
The equation is given as;
2HCl(aq) + Zn(s) + H2(g) + ZnCl2(aq)
In writing an ionic equation, only the aqueous compounds dissociates into ions. This means HCl and ZnCl2 would dissociate to form ions.
This is given as;
2H+ + 2Cl- + Zn(s) --> H2(g) + Zn2+ + 2Cl-
The correct option is;
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
When electrons are filling energy levels, the lowest energy sublevels are occupied first. This is Hund's rule.
Hund's rules state that:
Every orbital in a sublevel has to be singularly occupied before any other orbital is able to be doubly occupied.
All of the electrons in single occupied orbitals have to have the same spin to maximize the total spin.
Answer:
56
Explanation:
1 mole of gas at STP occupies 22.4 L of the gas
2.5 mole of the gas at STP occupies 22.4×2.5 L of the gas
so 2.5 mole of the gas at STP occupies 56 L of the gas .
When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
Learn more about balancing equation here:
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