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NNADVOKAT [17]
4 years ago
7

The formula for the carbonate ion is CO32−. Predict the formula for carbonic acid.

Chemistry
2 answers:
klio [65]4 years ago
7 0

Answer : The formula of carbonic acid is, H_2CO_3

Explanation :

Acid : It is a substance that has ability of donating proton or hydrogen ion, H^+. That means, in acid hydrogen should be present in the compound.

The element carbonate is the non-metal polyatomic ion. The stable oxidation state of phosphate ion (CO_3^{2-}) is, (-2).

When the hydrogen combine with the carbonate ion, it forms hydrogen carbonate (carbonic acid). The formula of carbonic acid is, H_2CO_3.

It is an ionic compound in which the a metal (hydrogen) react with the non-metal (carbonate ion) and combine with the an ionic bond by the criss-cross method.

Hence, the formula of carbonic acid is, H_2CO_3

Vikki [24]4 years ago
6 0
H2CO3 is the formula
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Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

So, for the reaction of 1.57 moles of NH_{3}, \Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ

As, \Delta G^{0} is negative therefore reaction is product favored under standard condition.

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A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
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Initial pressure                  3              1              0

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P = 4 - 3.75

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Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

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Pressure of O2 = 0.75 atm

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Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

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K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

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K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

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