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Nataliya [291]
3 years ago
8

ased on the thermodynamic properties provided for methanol (below and to the right), determine from the following list the equat

ion(s) necessary to calculate the amount of energy needed for 22.8 kg of methanol from all gas at 65°C to –112°C.
Chemistry
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

The necessary equations necessary are explained in each step in the solution.

Explanation:

For conversion of methanol gas at 65°C to -112°C

the equations would be as follows,

step 1 : gas at 65°C to liquid at 65°C

q = n(-37 kJ/mol)

step 2 : liquid at 65°C to liquid at -94°C

q = n(81.1 J/mol°C), where dT is negative

step 3 : liquid at -94°C to solid at -94°C

q = n(-3.18 kJ/mol)

step 4 : solid at -94°C to solid at -112°C

q = n(48.7 J/mol°C), where dT is negative

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Answer:

Tube 1: 10⁻¹, tube 2: 10⁻², tube 3: 10⁻³, tube 4: 10⁻⁴, tube 5: 10⁻⁵.

Explanation:

Serial dilutions are dilutions that the concentration decreases by the same quantity in each successive step. It means that the undiluted will be used for the first step, then the first will be used for the second, and successively. So, for a 10-fold, the concentration must decrease 1/10 in each step, it means that the dilution will be 1/10 in the first one (because it's 1 in tube 0).

In tube 1, the dilution is 1/10 = 0.1 = 10⁻¹;

In tube 2, the dilution will decrease more 1/10, so it will be 1/100x10 = 1/100 = 0.01 = 10⁻²;

In tube 3, it will be 1/1000x1/10 = 1/1000 = 0.001 = 10⁻³

In tube 4, it will be 10⁻⁴, and

In tube 5, it will be 10⁻⁵.

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3 years ago
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4 years ago
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guajiro [1.7K]

Answer: The equilibrium constant, K_c, for the reaction is 0.061.

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Equilibrium concentration of PCl_5 = \frac{31.8}{100}\times 0.039=0.012M  

The given balanced equilibrium reaction is,

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Initial conc.              0.039 M                0 M        0 M

At eqm. conc.     (0.039-x) M              (x) M   (x) M

Given : (0.039-x) = 0.012

x = 0.027

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[Cl_2]\times [PCl_3]}{[PCl_5]}

Now put all the given values in this expression, we get :

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Here, Na^{+} and NO_{3}^{-} ions are spectator ions as they remain present on both side of total ionic equation.

So, net ionic equation:

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