How do you write the complete ionic equation and net ionic equation for ammonium chloride and copper (II) sulfate in water?
If your aspirin sample has an absorbance of 0.500 in a 1.00 cm cell, what will be its absorbance in a 1.50 cm cell. What would be the percent change in the absorbance if you used the larger cell and didn’t factor in the change in path length? The equation is y = 0.148X-0.058
I found the absorbance, which was about 0.3139. I just don’t understand the second part of the question.
To write the complete ionic equation and net ionic equation for a reaction, you first need to know the reactants and products involved.
In this case, the reactants are ammonium chloride (NH4Cl) and copper (II) sulfate (CuSO4), both in water (H2O).
Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between ammonium chloride and copper (II) sulfate is:
2NH4Cl + CuSO4 -> 2NH4SO4 + CuCl2
Step 2: Write the complete ionic equation.
In the complete ionic equation, all soluble ionic compounds are split into their individual ions. Water (H2O) is left intact because it does not ionize.
Complete ionic equation:
2Na+ + 2Cl- + Cu2+ + SO42- -> 2Na+ + 2SO42- + Cu2+ + 2Cl-
Step 3: Write the net ionic equation.
The net ionic equation retains only the ions involved in the reaction and eliminates the spectator ions (ions that do not participate in the reaction and appear on both sides of the equation).
Net ionic equation:
2Cl- + Cu2+ -> CuCl2
In this case, both Na+ and SO42- ions are spectator ions since they appear on both sides of the equation. The final net ionic equation represents the core reaction occurring between the ammonium chloride and copper (II) sulfate in water.