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o-na [289]
2 years ago
12

Calculate the final temperature of 12.0 g of Argon (considered as an ideal gas) that is expanded reversibly and adiabatically fr

om V1 = 1.00 dm3 at T1 = 273.15 K to V2 = 3.00 dm3 . CV,m(Ar) = (3/2) * R.
Chemistry
1 answer:
Deffense [45]2 years ago
3 0

Answer:

Explanation:

For adiabatic change the expression is  

= PV^\gamma=constant

=(\frac{RT}{V})V^\gamma = constant

=TV^{\gamma-1}  = constant

=T_1V_1^{\gamma-1}  ==T_2V_2^{\gamma-1}

for Argon γ = 1.67

273.15 x 1 = T₂ x 3^{1.67-1}

T₂ = 273.15 / 3^{0.67}

= 273.15/ 2.0877

= 130.83 K.

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aleksley [76]

Answer : The value of ΔG expressed in terms of F is, -1 F

Explanation :

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E^o=(0.1V)-(-0.4V)=+0.5V

Now we have to calculate the standard cell potential.

Formula used :

\Delta G^o=-nFE^o

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\Delta G^o=-(2\times F\times 0.5)

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Therefore, the value of ΔG expressed in terms of F is, -1 F

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3 years ago
A chemical equation is shown below. KNO3 → KNO2 + O2 What are the coefficients that should be added to balance this equation? Us
uysha [10]

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Answer:

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Answer:

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