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3 years ago

Calculate the molality of a solution containing 22.75 g of glycerol (C3H8O3) in 79.6 g of ethanol (C2H5OH).

Chemistry

1 answer:

3241004551 [841]3 years ago

5 0

Answer:

3.11 mol/kg

Explanation:

Molality M = number of moles of solute, n/mass of solvent, m

To calculate the number of moles of glycerol (C₃H₈O₃) in 22.75 g of glycerol, we find its molar (molecular) mass, M',

So, M' = 3 × atomic mass of carbon + 8 × atomic mass hydrogen + 3 × atomic mass of oxygen

= 3 × 12 g/mol + 8 × 1 g/mol + 3 × 16 g/mol = 36 g/mol + 8 g/mol + 48 g/mol = 92 g/mol.

So, number of moles of glycerol, n = m'/M' where m' = mass of glycerol = 22.75 g and M' = molecular mass of glycerol = 92 g/mol

So, n = m'/M'

n = 22.75 g/92 g/mol

n = 0.247 mol

So, the molality of the solution M = n/m

Since m = mass of ethanol = 79.6 g = 0.0796 kg, substituting the value of n into the equation, we have

M = 0.247 mol/0.0796 kg

M = 3.11 mol/kg

So, the molality of the solution is 3.11 mol/kg.

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A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

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Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

$E^0_{[Ag^{+}/Ag]}=+0.80V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : $Cu(s)\rightarrow Cu^{2+}(aq)+2e^-$

Reduction half reaction (Cathode) : $Ag^{+}(aq)+e^-\rightarrow Ag(s)$

Thus the overall reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

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Answer;

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Explanation:

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2 years ago

describe how a pure dry sample of solid lead carbonate can be obtained from sodium carbonate solution and lead nitrate solution

djyliett [7]

Answer:

Soluble salts can be made by reacting acids with soluble or insoluble reactants. Titration must be used if the reactants are soluble. Insoluble salts are made by precipitation reactions.

Making insoluble salts

An insoluble salt can be prepared by reacting two suitable solutions together to form a precipitate.

Determining suitable solutions

All nitrates and all sodium salts are soluble. This means a given precipitate XY can be produced by mixing together solutions of:

X nitrate

sodium Y

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X = calcium and Y = carbonate

mix calcium nitrate solution and sodium carbonate solution together

calcium nitrate + sodium carbonate → sodium nitrate + calcium carbonate

Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3(aq) + CaCO3(s)

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2 years ago