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krek1111 [17]
2 years ago
14

15 g of gold and 25 g of silver are mixed to form a single-phase ideal solid solution.

Chemistry
1 answer:
Tatiana [17]2 years ago
8 0

Answer:

1. How many moles of solution are there. Ans: 0.3079193mol

2. Mole fraction for gold : 0.2473212

Mole fraction for silver: 0.7526787

3. Molar entropy of mixing for gold: 2.87285j/k

Molar entropy of mixing for silver: 1.77804j/k

4. Total entropy of mixing: 4.65089j/k

5. Molar free energy: -2325.445kj

6. Chemical potential for silver: -1750.31129j/mol

Chemical potential for gold: -575.13185j/mol

Explanation:

(1)

molar mass of silver = 107.8682g/mol

Molar mass of gold= 196.96657g/mol

Therefore mole = mass/molar mass

For silver: 25g/107.8682g/mol = 0.2317643mol

For gold: 15g/196.96657g/mol= 0.076155mol

Total number of mole= 0.2317643+0.076155= 0.30791193mol

(2)

Mole fraction for silver= 0.2317643/0.3079193= 0.7526787

Mole fraction for gold=0.076155/0.3079193=0.2473212

(3)

The molar entropy mixing ∆Sm= -RXi×lnXi

R= gas constant= 8.3144598

Xi = mole fraction

For silver:

-8.3144598×0.7526787( ln0.7526787)= 1.77804j/k

For gold:

-8.3144598×0.2473212( ln0.2473212)= 2.87285k/j

(4)

Total entropy= 1.77804+2.87285=4.65089k/j

(5)

Molar free energy change at 500°C

G=H-TS

Where G= Gibbs free energy

H= enthalpy,. T= Temperature, S= entropy

H=0, T=500+ 273=773k, S=4.65089

Therefore

G= 0- 773x4.65089= -3595.138kj

(6)

Chemical potential = Gibbs free energy × mole fraction

For silver:

-3595.138×0.7526787=-2705.9837j/mol

For gold:

-3595.138×0.2473212= -889.15381j/mol

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vredina [299]

Answer:

volume in L = 0.25 L

Explanation:

Given data:

Mass of Cu(NO₃)₂ = 2.43 g

Volume of KI = ?

Solution:

Balanced chemical equation:

2Cu(NO₃)₂  + 4KI    →    2CuI + I₂ + 4KNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

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Number of moles = 0.013 mol

Now we will compare the moles of Cu(NO₃)₂ with KI.

                        Cu(NO₃)₂       :              KI    

                              2              :               4

                            0.013          :            4 × 0.013=0.052 mol

Volume of KI:

<em>Molarity = moles of solute / volume in L</em>

volume in L = moles of solute /Molarity

volume in L =  0.052 mol / 0.209 mol/L

volume in L = 0.25 L

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Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

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<span><span>
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<span><span>Hope this answers the question. Have a nice day.</span></span>

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