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WITCHER [35]
2 years ago
14

When salt is added to water, the entropy, S,

Chemistry
1 answer:
Dovator [93]2 years ago
5 0

Answer:

C. number of particles

Explanation:

Entropy is the measure of disorderliness of a system. Remember that when you dissolve salt in water, you increase the number of particles in the solution. The greater the number of particles in solution, the greater the entropy of the  solution system.

Hence dissolution of a salt in water increases the entropy by increasing the number of particles in solution leading to the inequality; Ssolution > Swater + Ssalt.

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The heat of fusion AH, of ethyl acetate (C4H802) is 10.5 kinol. Calculate the change in entropy as when 398. g of ethy, acetate
Hitman42 [59]

<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

\text{Moles of ethyl acetate}=\frac{398g}{88.11g/mol}=4.52mol

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{fusion}}{T}

where,  

\Delta S = Entropy change  = ?

n = moles of ethyl acetate = 4.52 moles

\Delta H_{fusion} = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol   (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = 84.0^oC=[84+273]K=357K

Putting values in above equation, we get:

\Delta S=\frac{4.52mol\times 10500J/mol}{357K}\\\\\Delta S=132.9J/K

Hence, the entropy change of the ethyl acetate is 133. J/K

7 0
2 years ago
What is the mass of 2.25 x 10^25 atoms of lead
NemiM [27]

Answer: the amount of a substance that contains 6.02 x 1023 respective particles of that

substance

Avogadro’s number: 6.02 x 1023

Molar Mass: the mass of one mole of an element

CONVERSION FACTORS:

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

Try:

1. How many atoms are in 6.5 moles of zinc?

6.5 moles 6.02 x 1023 atoms = 3.9 x 1024 atoms

1 mole

2. How many moles of argon are in a sample containing 2.4 x 1024 atoms of argon?

2.4 x 1024 atoms of argon 1 mole = 4.0 mol

6.02 x 1023 atoms

3. How many moles are in 2.5g of lithium?

2.5 grams Li 1 mole = 0.36 mol

6.9 g

4. Find the mass of 4.8moles of iron.

4.8 moles 55.8 g = 267.84 g = 270g

1 mole

MOLE PARTICLES

(ATOM)

MASS

(g)

1 mole = molar

mass

(look it up on

the PT!)

1 mole =

6.02 x 1023

atoms

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

Two Step Problems:

1. What is the mass of 2.25 x 1025 atoms of lead?

2.25 x 1025 atoms of lead 1 mole 207.2g = 7744.19g = 7740g

6.02 x 1023 atoms 1 mole

2. How many atoms are in 10.0g of gold?

10 g gold 1 mole 6.02 x 1023 atoms = 3.06 x 1022 atoms

197.0g 1 mole

PRACTICE PROBLEMS:

1. How many moles are equal to 625g of copper?

625g of copper 1 mol = 9.77 mol Cu

64 g Cu

2. How many moles of barium are in a sample containing 4.25 x 1026 atoms of barium?

4.25 x 1026 atoms of barium 1 mol = 706 mol

6.02 x 1023 atoms

3. Convert 2.35 moles of carbon to atoms.

2.35 moles 6.02 x 1023 atoms = 1.41 x 1024 atoms

1 mole

4. How many atoms are in 4.0g of potassium?

4.0 g 1 mole 6.02 x 1023 atoms = 6.2 x 1022 atoms

39.1 g 1 mole

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)

5. Convert 9500g of iron to number of atoms in the sample.

9500 g Fe 1 mole 6.02 x 1023 atoms = 1.0 x 1026 atoms

55.8g 1 mole

6. What is the mass of 0.250 moles of aluminum?

0.250 moles 27.0g = 6.75 g Al

1 mole

7. How many grams is equal to 3.48 x 1022 atoms of tin?

3.48 x 1022 atoms 1 mole 118.7g = 6.86 g Sn

6.02 x 1023 atoms 1 mole

8. What is the mass of 4.48 x 1021 atoms of magnesium?

4.48 x 1021 atoms 1 mole 24.3g = 0.181 g Mg or 1.8 x 10-1

6.02 x 1023 atoms 1 mole

9. How many moles is 2.50kg of lead?

2.50kg 1000 g 1 mol = 12.1 mol

1 kg 207.2 g

10.Find the mass, in cg, of 3.25 x 1021 atoms of lithium.

3.25 x 1021 atoms 1 mol 6.9g 100cg = 3.7

Explanation:

5 0
2 years ago
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