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Nataly_w [17]
3 years ago
9

liv sheldon Given the balanced equation for an organic reaction: C2H2 + 2Cl2 → C2H2Cl4 This reaction is best classified as *

Chemistry
1 answer:
Afina-wow [57]3 years ago
4 0

Answer:

An addition reaction

Step-by-step explanation:

In an addition reaction, two or more molecules come together to form a single product, for example,

C₂H₂ + 2Cl₂ ⟶ C₂H₂Cl₄

This reaction consists of two successive additions. The product of the first reaction becomes a reactant and adds a second molecule of Cl₂ to form C₂H₂Cl₄

    C₂H₂ + Cl₂ ⟶ <em>C₂H₂Cl₂ </em>

<em><u>C₂H₂Cl₂</u></em><u> + Cl₂ ⟶ C₂H₂Cl₄ </u>

 C₂H₂ + 2Cl₂ ⟶ C₂H₂Cl₄

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Answer:

The forces involved in a collision are equal in size and directed in the opposite direction, and they accelerate both objects. Each object accelerates equally in collisions with things of equal mass.

Explanation:

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An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the
victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

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pH + pOH = 14

We are given:

pH = 8.08

pOH=14-8.08=5.92

  • To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

Putting values in above equation, we get:

5.92=-\log[OH^-]

[OH^-]=10^{-5.92}=1.202\times 10^{-6}M

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of X^- ions follows:

                    X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b

<u>Initial:</u>              0.1

<u>At eqllm:</u>        0.1-x                           x              x

Concentration of OH^-=x=1.202\times 10^{-6}M

The expression of K_b for above equation follows:

K_b=\frac{[OH^-][HX]}{[X^-]}

Putting values in above expression, we get:

K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M

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K_w=K_b\times K_a

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K_w = Ionic product of water = 10^{-14}

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K_b = Base dissociation constant = 1.445\times 10^{-11}

Putting values in above equation, we get:

10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}

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K_a\text{ for HF}=6.8\times 10^{-6}

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