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Yanka [14]
3 years ago
7

Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation.CH3COO-+HCO

3-CH3COOH+CO32-Brønsted-LowryBrønsted-LowryBrønsted-LowryBrønsted-LowryIn this reaction:The formula for the conjugateof CH3COO- isThe formula for the conjugateof HCO3- is
Chemistry
1 answer:
NARA [144]3 years ago
7 0

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor

The given equilibrium reaction is,

CH_3COO^-+HCO_3^-\rightleftharpoons CH_3COOH+CO_3^{2-}

In this reaction, CH_3COO^- and HCO_3^- are act as Bronsted Lowry base and acid respectively and CH_3COOH and CO_3^{2-} are act as Bronsted Lowry acid and base respectively.

The formula of conjugate of CH_3COO^- is, CH_3COOH

The formula of conjugate of HCO_3^- is, CO_3^{2-}

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<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>

Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where, 
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?

By applying the formula,
103591 Pa x  </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
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</span>Moles (mol) = mass (g) / molar mass (g/mol)<span>

Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
   molar mass of the gas = mass / moles
                                          = 0.281 g / </span>3.90 x 10⁻³ mol
<span>                                          = 72.05 g/mol

</span>
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