What is the Kb expression for aniline (C6H5NH2)?

2 answers:

4 0

This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]

hope this helps:)

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navik [9.2K]3 years ago

3 0

the answer is A for #platofam

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A solution contains an unknown amount of dissolved magnesium. Addition of

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<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Mg²⁺: 1 mole
Na₂CO₃: 1 mole
MgCO₃: 1 mole
Na⁺: 2 moles

The molar mass of the compounds is:

Mg²⁺: 24.3 g/mole
Na₂CO₃: 106 g/mole
MgCO₃: 84.3 g/mole
Na⁺: 23 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
Na₂CO₃: 1 mole ×106 g/mole= 106 grams
MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
Na⁺: 2 moles ×23 g/mole= 46 grams

<h3>Mass of magnesium dissolved</h3>

The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?

$mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}$