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OLEGan [10]
1 year ago
14

Derive third law of thermodynamics equation

Chemistry
1 answer:
Hoochie [10]1 year ago
3 0

Answer:

The Third Law of thermodynamics states that the entropy of a pure substance in a perfect crystalline state at zero temperature is zero.

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What is the silver ion concentration in a solution prepared by mixing 425 mL 0.397 M silver nitrate with 427 mL 0.459 M sodium p
Lisa [10]

Answer:

0 M is the silver ion concentration in a solution prepared mixing both the solutions.

Explanation:

molarity=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}

Moles of silver nitrate = n

Volume of the solution = 425 mL = 0.425  L (1 mL = 0.001 L)

Molarity of the silver nitrate solution = 0.397 M

n=0.397 M\times 0.425 L=0.1687 mol

Moles of sodium phosphate = n'

Volume of the sodium phosphate solution = 427 mL = 0.427  L (1 mL = 0.001 L)

Molarity of the sodium phosphate solution = 0.459 M

n'=0.459 M\times 0.427 L=0.1960 mol

3AgNO_3+Na_3PO_4\rightarrow Ag_3PO_4+3NaNO_3

According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :

\frac{1}{3}\times 0.1687 mol=0.05623 mol of sodium phosphate

This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.

So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.

0 M is the silver ion concentration in a solution prepared mixing both the solutions.

4 0
3 years ago
Some animals only have _______ cells
White raven [17]

Answer:

Animals can have up to 40 trillion cells.

An amoeba (type of animal) has just one cell.

3 0
2 years ago
Read 2 more answers
10) Air masses are different in many ways. Which of these is NOT different?
grandymaker [24]

Answer:D. Ratio of oxygen/nitrogen

Explanation: the ratio will never change no matter the air pressure!

7 0
2 years ago
Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota
Ghella [55]

Answer:

(a) ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = -2.881 J/K; total change in entropy = 0

(b)ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = 0 ; total change in entropy = 2.881 J/K

(c) ΔS_{sys}  = 0 ; ΔS_{sur}  = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = V_{1}

Final volume = V_{2} = 2V_{1}

(a) Change in entropy of the system ΔS_{sys} = nRIn\frac{V_{2} }{V_{1} }

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

ΔS_{sys} = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding ΔS_{sur} = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, ΔS_{sys}  = 2.881 J/K

Since surrounding does not change in this process ΔS_{sur} = 0.

total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

ΔS_{sys}  = 0

Since heat energy is not transferred from the system to the surrounding

ΔS_{sur}  = 0

total change in entropy = ΔS_{sys}+ΔS_{sur} = 0

6 0
2 years ago
When you burned copper, the product copper
MatroZZZ [7]

Answer:

Oxygen.

Explanation:

The copper must be combined with something in the air.

6 0
2 years ago
Read 2 more answers
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