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2 years ago

A- is a weak base. which equilibrium corresponds to the equilibrium constant ka for ha?

Chemistry

1 answer:

vekshin12 years ago

5 0

Answer:

$K_a=\frac{[H^+][A^-]}{[HA]}$

Explanation:

ka is defined as the dissociation constant of an acid. It is defined as the ratio of concentration of products to the concentration of reactants.

For the dissociation of weak acid, the chemical equation follows:

$HA\rightleftharpoons H^++A^-$

The equilibrium constant is defined by the equilibrium concentration of products over reactants:

$K_a=\frac{[H^+][A^-]}{[HA]}$

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Answer:

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Explanation:

Oxidation reactions are defined as,

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(ii) Addition of Oxygen:

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3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

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2 years ago

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beks73 [17]

Reactivity is a chemical property of a substance. According to EPA regulations, it is normally unstable and readily

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2 years ago

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Answer is
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2 years ago

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