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tatyana61 [14]
3 years ago
6

152 dm^3 of gas has a pressure of 98.6 kpa. at what pressure will the volume be quartered

Chemistry
1 answer:
djverab [1.8K]3 years ago
6 0

Answer:

P₂ = 394.4 KPa

Explanation:

Given data:

Volume of gas = 152 dm³

Pressure of gas = 98.6 KPa

Final pressure = ?

Final volume = quartered = 1/4×152 = 38 dm³

Solution:

P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

P₂ = 98.6 KPa .  152 dm³ / 38 dm³

P₂ = 14987.2 KPa.  dm³ / 38 dm³

P₂ = 394.4 KPa

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tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)

The half-cell reactions are:

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First we have to calculate the cell potential for this reaction.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

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Now put all the given values in the above equation, we get:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

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3 years ago
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Mariulka [41]

Answer:

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<em>Note: The question is incomplete. The complete question is given as follows:</em>

<em>A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm. </em>

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Explanation:

Using the formula of heat, Q = mc∆T  

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When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water

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c = 0.567 J/g.°C

Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.

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