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Paha777 [63]
3 years ago
7

What is a displacement reactions​

Chemistry
2 answers:
Alisiya [41]3 years ago
6 0

<h2><u>Question</u><u>:</u>-</h2>

\textsf{what is a displacement reactions?   }      

<h2><u>Answer</u><u>:</u></h2>

\large \underbrace{ \underline{ \sf Displacement \: reaction}}

The reaction in which more reactive element replace a less reactive element from its compound is called displacement reaction.

Ex:-

\bold{\large Zn+HCl \longrightarrow ZnCl_2+H_2}

jek_recluse [69]3 years ago
3 0
<h3> Displacement Reaction:</h3>

A displacement reaction is the one wherein the atom or a set of atoms is displaced by another atom in a molecule. For instance, when iron is added to a copper sulphate solution, it displaces the copper metal.

A + BC → AC + B

The above equation exists when A is more reactive than B.

A and B have to be either:

Halogens where C indicates a cation.

Different metals wherein C indicates an anion.

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Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

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Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

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