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forsale [732]
2 years ago
15

A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a SoaveRedlich-Kwong gas and with

C ∗ P = 100 J/(mol·K), from 300 K and 2 m3 to 800 K and 0.02 m3 by using less than 5 MJ of work. Is this possible?
Chemistry
1 answer:
Setler79 [48]2 years ago
3 0

Explanation:

The given data is as follows.

    Moles of propylene = 100 moles,    T_{i} = 300 K

    T_{f} = 800 K,    V_{i} = 2 m^{3}

    V_{f} = 0.02 m^{3},   C_{p} of propylene = 100 J/mol

Now, we assume the following assumptions:

Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.

            W = mC_{p} \Delta T

     100 moles \times 100 J/mol K (800 - 300) K

                 = 5 \times 10^{6} J

                 = 5 MJ

Thus, we can conclude that a minimum of 5 MJ work is required without any friction.

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Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh
Naya [18.7K]

Answer:

1.82 g   is the maximum mass of CuS.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

<u>For CuCl_2 : </u>

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of CuCl_2 :

Moles=0.500 \times {38.0\times 10^{-3}}\ moles

<u>Moles of CuCl_2  = 0.019 moles </u>

<u>For (NH_4)_2S : </u>

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of (NH_4)_2S :

Moles=0.600 \times {42.0\times 10^{-3}}\ moles

<u>Moles of (NH_4)_2S  = 0.0252 moles </u>

According to the given reaction:

CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}

1 mole of CuCl_2 reacts with 1 mole of (NH_4)_2S

So,  

0.019 mole of CuCl_2 reacts with 0.019 mole of (NH_4)_2S

Moles of (NH_4)_2S = 0.019 mole

Available moles of (NH_4)_2S = 0.0252 mole

<u>Limiting reagent is the one which is present in small amount. Thus, CuCl_2 is limiting reagent.</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of CuCl_2 gives 1 mole of CuS

0.019 mole of CuCl_2 gives 0.019 mole of CuS

Moles of CuS formed = 0.019 moles

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

<u>1.82 g   is the maximum mass of CuS.</u>

5 0
2 years ago
trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is
spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 cm^{-1}.

Explanation:

Bands that appear above 3000 cm^{-1}  are often unsaturation diagnoses suggest. The band at 3000- 3100 cm^{-1} is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

4 0
2 years ago
1. Which liquid sample is a pure substance?
IRISSAK [1]

The whole Activity , poem and paragraph is missing in the question.

Answer:

(1) Liquid A

(2) Solid A

Explanation:

Using this part of the given poem

Substances and mixtures behave differently,

During boiling and melting most especially

Boiling point of substance is fixed while mixture is not

Substance melts completely but mixture does not

The boiling point of the Pure substance remain fixed after reaching its boiling point this is shown by Liquid A

Solid A is melting completely so Solid A is a pure substance.

6 0
2 years ago
How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?
mash [69]

Answer : Both solutions contain 3.011 X 10^{23} molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain 3.011 x 10^{23} molecules.

Avogadro's Number is  N_{A} =  6.022 X 10^{23} which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules  = N_{A} X M;

∴  Number of molecules =  6.022 X 10^{23} X 0.5 mol/L X 1 L which will be  = 3.011 X 10^{23}

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

7 0
2 years ago
Read 2 more answers
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
2 years ago
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