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katen-ka-za [31]
1 year ago
14

Complete the charge balance equation for an aqueous solution of h2co3 that ionizes to hco−3 and co2−3.

Chemistry
1 answer:
Zielflug [23.3K]1 year ago
8 0

The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The equation for aqueous solution of H₂CO₃ is

H₂CO₃ → H₂O + CO₂

The charge balance equation is

[HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] =  2[CO₃⁻²] + [H⁺] + [OH⁻]

Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
#SPJ4

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The entropy change for a real, irreversible process is equal to:______.
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Answer:

The entropy change for a real, irreversible process is equal to <u>zero.</u>

The correct option is<u> 'c'.</u>

Explanation:

<u>Lets look around all the given options -:</u>

(a)  the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.

(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .

(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.

(d) impossible to tell. This option is invalid , thus incorrect .

<u>Hence , the correct option is 'c' that is zero.</u>

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3 years ago
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Answer: There are 4.45\times 10^{-4}moles of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

Explanation:

According to ideal gas equation:

PV=nRT

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V = Volume of gas = 9.55 ml = 0.00955 L   (1 L=1000ml)

n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =35^0C=(35+273)K=308K

n=\frac{PV}{RT}

n=\frac{1.18atm\times 0.00955L}{0.0821L atm/K mol\times 308K}=4.46\times 10^{-4}moles

Thus there are 4.45\times 10^{-4}moles of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

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