Um, I think it’s: k is potassium and F is fluorine so potassium Fluoride
Answer:
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
Explanation:
The balanced reaction between Na2CO3 and HCl is given as;
Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)
The next step is o express the species as ions.
The complete ionic equation for the above reaction would be;
2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + CO₂ (g) + H₂O (l)
The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.
The spectator ions are; Na⁺ and Cl⁻
The net ionic equation is given as;
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
In nitration u add two concentration acids HNO3 and H2SO4 both are strong acids. always the acid base neutralization reactions are fastest. even though NH2 group in aniline is a weak base, its neutralization with a very strong acid is still faster than nitration. hence if u carry the acid base reaction u would end up with an NH3+ group. hence the meta substitution can't be rule out here and u would get all products ortho,meta and para...
Hopefully this helps sorry
Given reaction is NO(g)+1/2 O₂(g)→NO₂(g)
ΔGf⁰ [NO₂(g)] = 51.9 kJ/mol
ΔGf⁰ [NO(g)] = 86.6 kJ/mol
ΔGf⁰ <span>[O2(g)] = 0 kJ/mol</span>
ΔG⁰rxn = ΔGf⁰(products) - ΔGf⁰( reactants)
= ΔGf⁰ [NO₂(g)] - {ΔGf⁰ [NO(g)]+ (1/2) ΔGf⁰ [O₂(g)]}
= 51.9 kJ/mol - { 86.6 kJ/mol + 0}
= -34.7 kJ/mol
Therefore, ΔG⁰rxn = -34.7 kJ/mol
