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777dan777 [17]
3 years ago
10

2.0 kilograms of lithium hydroxide can “scrub”, or remove, how many liters of carbon dioxide at standard conditions?

Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

0.94 L

Explanation:

The equation of the reaction is;

2LiOH + CO2 -----> Li2CO3 + H2O

Number of moles LiOH = 2.0/23.95 g/mol = 0.084 moles

2 moles of LiOH scrubs 22.4 L of CO2

0.084 moles of LiOH scrubs 0.084 moles × 22.4/2

= 0.94 L

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Quinine (C20H24N2O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine
dolphi86 [110]

Answer:

The pH of saturated solution of the quinine is 10.05

Explanation:

Quinine (Q) is C20H24N2O2 has a molar mass of 324.4 g/mol

Q can behave as a weak base. Kb and pKb can be calculated for weak bases

pKb1 is to be considered when solving the question.

pKb1 = 5.1

Step 1 : Calculate the Kb of Quinine

            pKb1 = - log [kb]

                5.1  = - log [kb]

                take Antilog of both side

             [kb] = 7.94 x 10∧-6

Step 2: Calculate the concentration of saturated solution of Q in mol/dm3

           From the question, 1900 ml of solution contains 1 g of Q

           Therefore,  1000 ml of solution will contain........... x g of Q

           x = 1000 /1900

           x = 0.526 g in 1 dm3

In calculating concentration in mol/dm3,

Concentration in mol/dm3 = concentration in g /dm3 divided by molar mass

Molar mass of Q = 324.4

Concentration in mol/dm = 0.526 /324.4

                                         = 0.0016 mol/dm3

Step 3: Calculating the Concentration of OH-

            At Equilibrium, Kb = x² / 0.0016

            7.94 x 10∧-6 = x² / 0.0016

            x = √ 0.0016 × 7.94 x 10∧-6

            x = 1. 127 × 10∧-4 mol/dm3

The concentration of OH- = 1. 127 × 10∧-4 mol/dm

Step 4:  Calculating the pH of Quinine

           Recall, pOH = - log [OH-]

           pOH = - log [1. 127 × 10∧-4]

           pOH = 3.948

           Also recall that pH + pOH = 14

           pH = 14 - 3.948

           pH = 10.05

           

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