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m_a_m_a [10]
1 year ago
12

PLEASE HELP CHEMISTRY

Chemistry
1 answer:
eduard1 year ago
7 0

When sodium carbonate is dissolved in water, the equation is Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq).

When carbon dioxide is placed in water, aqueous carbon dioxide is formed:  CO_2 (g) --- > CO_2(aq)

<h3>Dissolution of compounds in water</h3>

Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.

Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.

While in solution, sodium carbonate dissociates into its component ions according to the following equation:

Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq)

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.

Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: CO_2 (g) --- > CO_2(aq)

More on the dissolution of substances can be found here: brainly.com/question/28580758

#SPJ1

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The given balanced equilibrium reaction is,

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Initial conc.              0.450 M               0        0

At eqm. conc.    (0.450 -2x) M         (2x) M   (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[O_2][SO_2]^2}{[SO_3]^2}

K_c=\frac{x\times (2x)^2}{0.450-2x)^2}

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K_c=\frac{0.055\times (2\times 0.055)^2}{0.450-2\times 0.055)^2}

K_c=0.0057

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