Answer:
H+(aq) + OH-(aq) ⟶H2O(l)
Explanation:
Step 1: The balanced equation
HCN(aq) + KOH(aq) ⟶ H2O (l) + KCN (aq)
H+(aq) + CN-(aq) + K+(aq) + OH-(aq) ⟶H2O(l) + K+(aq) + CN-(aq)
Step 2: The net ionic equation
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:
H+(aq) + OH-(aq) ⟶H2O(l)
Can u try putting this in english please
The chemical reaction of magnesium and sodium hydroxide would yield magnesium hydroxide and sodium. The chemical reaction is expressed as:
Mg(s)+ 2NaOH(aq)→Mg(OH)2(s)+2Na
In ionic form,
Mg(s) + 2Na+ + 2OH−(aq)→Mg(OH)2(s) + 2Na
Answer:
(a) -49.9 kJ/mol;
(b) To the right;
(c) 34.6 kJ/mol
Explanation:
(a) For this reaction, since it's at equilibrium and standard states, we know that we can apply the equation:

Substituting the given variables:

(b) Notice that this reaction is spontaneous, since
. This means reaction spontaneously proceeds to the right side. Besides, K > 1, this means products dominate over reactants, so reaction proceeds to the right.
(c) Given the expression of the formation constant, we can use the same expression to calculate the reaction quotient at non-standard conditions:
![Q_f = \frac{[Ni(NH_3)_6]^{2+}}{[Ni^{2+}][NH_3]^6} = \frac{0.010}{0.0010\cdot 0.0050^6} = 6.4\cdot 10^{14}](https://tex.z-dn.net/?f=Q_f%20%3D%20%5Cfrac%7B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%7D%7B%5BNi%5E%7B2%2B%7D%5D%5BNH_3%5D%5E6%7D%20%3D%20%5Cfrac%7B0.010%7D%7B0.0010%5Ccdot%200.0050%5E6%7D%20%3D%206.4%5Ccdot%2010%5E%7B14%7D)
Now, notice that
. In this case, we have an excess of the products, this means reaction will shift to the let left to restore the equilibrium.
Calculate:

Put the metal with the non-metal and use the charges to figure it out