Exploring Reactivity Series: A Chemical Interaction Study

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Zn (s) + Cu(NO3)2 (aq) → Cu(s) + Zn(NO3)2 (aq) (Eq. 2)

Regarding, the copper and sodium chloride, no visible change occurred after 5 minutes had passed. The copper metal remained at the bottom of the test tube undeterred by the NaCl. There may not have been a reaction because copper is lower on the reactivity series than sodium.

Metathesis: Precipitation (Nonredox)
The reaction between and KCl is a double replacement reaction because the cations Cu2+ and K+ switched places and formed two new compounds, copper (II) chloride and potassium nitrate. However, there was no observed precipitation in the resulting solution as both are soluble.

Balanced formula- Cu(NO3)2(aq) + 2KCl(aq) → CuCl2(aq) + 2KNO3(aq) (Eq. 3)
Total ionic- Cu2+(aq) + 2NO3-(aq) …show more content…

As 10 mL of NaOH was added drop wise, the solution began to have a pink tint. The fuchsia color did not permanently stay until 9mL of NaOH was added. The pH indicator demonstrated when exactly the reaction has neutralized. Because an acid and a base reacted to form water and a salt, the reaction is a neutralization …show more content…

9)
Total ionic- Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l) (Eq. 10)
Net ionic- NaOH(s) + H+ (aq) → H2O(l) + Na+ (aq) (Eq. 11)

Solid sodium hydrogen carbonate and HCl reacted violently, bubbling and foaming at a rapid pace. NaHCO3, baking soda in layman’s terms, is basic so it reacts strongly with acids such as HCL.

Balanced formula- NaHCO3(s) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) (Eq. 12)
Total ionic- H+(aq) + Cl-(aq) + Na+(aq) + HCO3-(aq) → Na+(aq) + Cl-(aq) + CO2(g) + H2O(l) (Eq. 13)
Net ionic- HCO3-(aq) + H+(aq) → H2O(l) + CO2(g) (Eq. 14)

Metathesis- Amphoteric Hydroxides and Complex Ion Formation
A diluted NaOH solution was created by mixing 6 drops of 4.0 M NaOH and 44 drops of distilled water. 20 drops is equivalent to 1 mL of liquid[E] so the new molarity of the solution can be calculated using an equation used for dilution calculations:

M1V1 = M2V2[F] (Eq. 15)
(4.0 mol/L)(0.3 mL) = (M2)(2.5 mL)
M2 = 0.48 M

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