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12345 [234]
3 years ago
9

Determine the pH of a 0.530 M solution of carbonic acid that has an acid dissociation constant of 4.4 x

Chemistry
2 answers:
Veronika [31]3 years ago
5 0

Answer:

Explanation:

H2CO3  =  H+   +  HCO3-   H2CO3  =  0.530 M  Ka =4.4 X10^-7

Ka =[H+][A-]/[HA]        [H+] = [A-]

0.530 M x 4.4 X10^-7=  [H+}^2

2.32 X10^-7  =  [H+}^2

23.2 X10^-8 = [H+}^2

4.83 X 10^-4 = [H+]

pH = - log 4.83 X 10^-4

pH = -(.68-4)

pH =-(-3.32)

pH= 3.32

Effectus [21]3 years ago
4 0

Answer:

it is D

Explanation:

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The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of H_2 gas and 1 mole of Br_2 liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

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Divide the equation by 2.

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

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We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.

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To ascertain the oxidation state, we have to comply with some rules:

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