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Murrr4er [49]
3 years ago
15

Barium hypochlorite and perchloric acid net ionic equation.

Chemistry
1 answer:
Lapatulllka [165]3 years ago
5 0

This problem asks for the net ionic equation between barium hypochlorite and perchloric acid. The answer then turns out to be  ClO^-(aq)+H^+(aq)\rightleftharpoons HClO(aq).

<h3>Net ionic equations</h3>

In chemistry, when a chemical reaction takes place, one can study three versions of the chemical equation representing it:

1. Complete molecular equation: shows all the substances, their phases of matter and every balanced atom. In this case, one can write:

Ba(ClO)_2(aq)+2HClO_4(aq)\rightarrow 2HClO(aq)+Ba(ClO_4)_2(aq)

2. Complete ionic equation: shows all the aqueous salts and strong both acids and bases split into cation and anion; however, weak both acids and bases, such as the hypochlorous acid here, remain unionized as well as solid precipitates, liquids or gases:

Ba^{2+}(aq)+2(ClO)^-(aq)+2H^+(aq)+2(ClO_4)^-(aq)\rightarrow 2HClO(aq)+Ba^{2+}(aq)+2(ClO_4)^-(aq)

3. Net ionic equation: cancels out the spectator ions (those equal on both reactants and products) and shows the relevant ions that undergo a reaction. In this case, both perchlorate and barium ions act as the spectator ones, consequently, we obtain:

2(ClO)^-(aq)+2H^+(aq)\rightarrow 2HClO(aq)\\\\(ClO)^-(aq)+H^+(aq)\rightarrow HClO(aq)

That can also be written at equilibrium due to the concept of weak acid for the formation of hypochlorous acid:

ClO^-(aq)+H^+(aq)\rightleftharpoons HClO(aq)

Learn more about spectator ions: brainly.com/question/15365313

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Answer:

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Explanation:

As we know

Q = m * c * change in temperature

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Substituting the given values we get  -

5750 = 335 * 4.2 * (X - 35.5)

X = 39.58 degree Celsius

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Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

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Explanation:

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4 years ago
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