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kupik [55]
1 year ago
14

A system fitted with a piston expands when it absorbs 52.7J of heat from the surroundings. The piston is working against a press

ure of 0.693 atm. the final volume is 63.2L. what was the initial volume of the system if the internal energy of the system decreased by 107.4J?
Chemistry
1 answer:
Allushta [10]1 year ago
6 0

Answer: The initial volume of the system is 60.29 L.

Explanation:

According to the first law of thermodynamics,

      \Delta U = Q - W

As it is given that heat is being added to the system so, \Delta H will be positive. And, work done on the system is negative and work done by the system is positive.

So here, \Delta U = -107.4 J

                    Q = 52.7 J

                    P = 0.693 atm

And,     W = PdV

or,        W = P (V_{final} - V_{initial})

So,       \Delta U = Q - P (V_{final} - V_{initial})

            -107.4 J = 52.7 J - 0.693 \times 101.325 (63.2 - V)

            -160.1 = -43.79 - 70.21 (63.2 - V)

           63.2 - V = \frac{160.1 + 43.79}{70.21}

                  - V =  2.90 - 63.2

                     V = 60.29 L

Thus, we can conclude that the initial volume of the system is 60.29 L.

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Answer:

See explanation

Explanation:

A flammable solvent refers to a solvent that catches fire easily. The precautions to be taken when working with flammable solvents are;

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5 0
2 years ago
A heat energy of 645 J is applied to a sample of glass with a mass of 28.4 g. Its temperature increases from –11.6 ∞C to 15.5 ∞C
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The heat that is required to raise the temperature of an object is calculated through the equation,
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Specific heat is therefore calculated through the equation below,
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Substituting,
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The value of specific heat from above equation is 0.838 J/g°C. 
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2 years ago
Use the thermodynamic data at 298 k below to determine the ksp for barium carbonate, baco3 at this temperature. substance: ba2+(
jeka57 [31]

Answer : the correct answer for ksp = 1.59 * 10⁻⁹

Following are the steps to calculate the ksp of reaction

BaCO₃ →Ba ²⁺ + CO₃²⁻ :

Step 1 : To find ΔG° of reaction :

ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :

ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .

Given : ΔG° for Ba²⁺ ( product )= -560.7 \frac{KJ}{Mol}

ΔG° for CO₃²⁻ (product ) =- 528.1 \frac{KJ}{Mol}

ΔG° BaCO₃ ( reactant) = –1139 \frac{KJ}{Mol}

Plugging value in formula :

ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )

⁻ = ( -560.7 \frac{KJ}{Mol} + 528.1 \frac{KJ}{Mol} ) - ( -1139 \frac{KJ}{Mol} )

= ( -1088.8 \frac{KJ}{Mol}) - (-1139 \frac{KJ}{Mol} )

= - 1088.8 \frac{KJ}{Mol} + 1139 \frac{KJ}{Mol}

ΔG° of reaction = 50.2 \frac{KJ}{Mol}

Step 2: To calculate ksp from ΔG° of reaction .

The relation between Ksp and ΔG° is given as :

ΔG° = -RT ln ksp

Where ΔG° = Gibb's Free energy R = gas constant T = Temperature

Ksp = Solubility constant product .

Given : ΔG° of reaction = 50.2 \frac{KJ}{Mol}

T = 298 K R = 8.314 \frac{J}{Mol * K}

Plugging values in formula

50.2 \frac{KJ}{mol}  =  -  8.314 \frac{J}{mol * K} * 298 K * ln  ksp

50.2 \frac{KJ}{mol}  =  - 2477.572 \frac{J}{mol} * ln K

((Converting 2477 \frac{J}{mol}  to \frac{KJ}{mol}

Since , 1 KJ = 1000 J So , 2477 \frac{J}{mol}  * \frac{1 KJ}{1000J}  = 2.477 \frac{KJ}{mol} ))

Dividing both side by - 2.477 \frac{KJ}{mol}

\frac{50.2\frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} = \frac{-2.477 \frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} * ln ksp

ln ksp = ln ksp = -20.27 \frac{KJ}{mol}

Removing ln :

ksp = 1. 59 * 10⁻⁹

8 0
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Licemer1 [7]

Answer:

a.)  

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Total:

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b.)  

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To melt the solid at -114°C:

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Total:

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Explanation:

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