Answer:
When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solution of sodium chloride. Write the conventional equation, total ionic equation, and net ionic equation for this reaction.
Explanation:
The word equation for the reaction is:
Copper (II) chloride(aq) + sodium carbonate (aq) ->sodium chloride (aq) + copper carbonate(s)
The balanced chemical equation of the reaction is:
The complete ionic equation is:
The net ionic equation is obtained from the complete ionic equation after removing the spectator ions:
Answer:
A. It's a 'clean' energy. Your cooch need a cleanup
Answer:
Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)
Step-by-step explanation:
Molecular Equation:
(NH₄)₂S(aq) + FeCl₂(aq) ⟶ 2NH₄Cl(aq) + FeS(s)
Ionic equation
:
2NH₄⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) ⟶ 2NH₄⁺(aq) + 2Cl⁻(aq) + FeS(s)
Net ionic equation
:
Cancel all ions that appear on both sides of the reaction arrow (underlined).
<u>2NH₄⁺(aq)</u> + S²⁻(aq) + Fe²⁺(aq) + <u>2Cl⁻(aq)</u> ⟶ <u>2NH₄⁺(aq) </u>+ 2<u>Cl⁻(aq) </u>+ FeS(s)
Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)
Answer:
The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl
Explanation:
Step 1: Data given
pH of a buffer = pKa + log ([A-]/[Ha])
a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)
pH = -log( 1.8 * 10^-5) + log (1/1)
pH = -log( 1.8 * 10^-5)
pH = 4.74
a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)
pH = -log( 4.9 * 10^-10) + log (1/1)
pH = -log( 1.8 * 10^-5)
pH = 9.30
a mixture of 1.0 M HCl and 1.0 M NaCl
The solution made from NaCl and HCl will NOT act as a buffer.
HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.
a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)
Ka * Kb = 1*10^-14
Ka = 10^-14 / 1.76*10^-5
Ka = 5.68*10^-10
pH = -log( 5.68*10^-10) + log (1/1)
pH = -log( 5.68*10^-10)
pH = 9.25
The most effective buffer at pH 9.25 will be a mixture of 1.0 M NH3 and 1.0 M NH4Cl
