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Ray Of Light [21]
2 years ago
9

Energy and Enthalpy Changes, Heat and Work -- Monatomic Ideal Gas dynamically generated plot 2.00-mol of a monatomic ideal gas g

oes from State A to State D via the path A→B→C→D: State A PA=11.0atm, VA=12.50L State B PB=11.0atm, VB=7.00L State C PC=25.0atm, VC=7.00L State D PD=25.0atm, VD=20.50L Assume that the external pressure is constant during each step and equals the final pressure of the gas for that step. Calculate q for this process. q = 1pts Tries 0/6 Calculate w for this process. w = -902 J Use w = -PΔV and sum over the individual steps. Check that you have the correct energy units and be sure you have the sign right (w is defined as the work done ON THE SYSTEM). 1pts Incorrect. Tries 4/6 Previous Tries Calculate ΔE for this process ΔE = 1pts Tries 0/6 Calculate ΔH for this process. ΔH = 1pts
Chemistry
1 answer:
larisa86 [58]2 years ago
6 0

(a) The heat generated in the process is 28 kJ.

(b) The work done in the process is determined as -28 kJ.

(c) The change in the internal energy is 0.

<h3>Heat of the isothermal compression </h3>

The heat generated in the process is negative done in the process.

W = -PΔV

W = -P(V₂ - V₁)

<h3>From A to B</h3>

W = -P(VB - VA)

W = -11(7 - 12.5)

W = 60.5 L.atm = 60.5 x 101.325 J/L.atm = 6,130.16 J

<h3>From C to D</h3>

W = -25(20.5 - 7)

W = -337.5 L.atm = -34,197.18 J

Total work , w = -34,197.18 J +  6,130.16 J = -28 kJ

q = - w

q = 28 kJ

<h3>Change in internal energy</h3>

ΔE = q + w

ΔE = 28 kJ - 28 kJ = 0

Learn more about change in internal energy here: brainly.com/question/17136958

#SPJ1

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For ethanol, propanol, and n-butanol the boiling points, surface tensions, and viscosities all increase. what is the reason for
Contact [7]
Moving from Ethanol through Propanol to Butanol the physical properties like boiling points, surface tension and viscosity increases because of the increases in intermolecular interactions between the molecules of given compounds.

Explanation:
                   Ethanol, propanol and butanol all have hydroxyl groups in common, means all have hydrogen bond intractions between their molecules. So, taking the hydrogen bonding interaction constant we are left with only the difference in the number of carbon atoms.
                    Butanol has the greatest physical properties than other two because it has four carbon atom chain. So, as we know the London Dispersion forces or Van der Waal forces increases with increase in molecular size and chain length of hydrocarbon.
                    Therefore, the strength of London forces is greater in butanol than other two while ethanol has the smallest chain comparatively hence, lowest physical properties.  
3 0
3 years ago
Ethanol (C2H5OH) and water (H2O) are both liquids at or near room temperature. When a sample of ethanol is poured into a beaker
BaLLatris [955]

Answer:

a. Are miscible because each can hydrogen bond with the other.

Explanation:

Both ethanol and water are miscible. The reason why they can both mix freely is due to the hydrogen bonds that will form between their molecular structure.

Hydrogen bonds are special dipole-dipole attraction between polar molecules in which hydrogen atoms are directly joined to an electronegative atom.

Ethanol has an hydroxyl group which will bond to form an intermolecular bond with the oxygen and hydrogen on the water molecule. This attraction makes them miscible.

5 0
3 years ago
enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t
laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM             |   [S]=KM                  |  [S]>>KM                     | Not true

____________  |   Half of the active  | Reaction rate is         | Increasing

[E_{free}] is about   |    sites are filled of  |    independent of      |  [E_{Total}] will                                            

 equal to [E_{total}]. |                                 |   [S]                             | lower KM

_____________________________________________|____________

[ES] is much       |                                 | Almost all active

 lower than         |                                 | sites are filled

[E_{free}]                  |                                 |

Explanation:

Generally the combined enzyme[ES] is mathematically represented as

                   [ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)

for Michaelis-Menten equation

Where [S] is the substrate concentration and K_M is the Michaelis constant

Considering the statement [S] < < K_M

  Looking at the equation [S] is denominator so it can be ignored(it is far too small compared to K_M)  hence the above equation becomes

               [ES] = \frac{[E_{total}][S]}{K_M}

Since [S] is less than K_M it means that \frac{[S]}{K_M}  < < 1

so it means that [ES] < < [E_{total}]

  What this means is that the  number of combined enzymes[ES] i.e the number of occupied site is very small compared to the the total sites [E_{total}]  i.e the total enzymes concentration which means that the free sites [E_{free}]  i.e the concentration of free enzymes is almost equal to [E_{total}]

Considering the second statement

      [S] = K_M

So  this means that equation one would now become

           [ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

      [S] >>K_M

In this case the K_M in the denominator of equation 1 would be neglected and the equation becomes

       [ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]

This means that almost all the sites are occupied with substrate

 The rate of this reaction is mathematically defined as

             v =\frac{V_{max}[S]}{K_M [S]}

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and V_{max}  is he maximum velocity of the reaction

In this case also the K_M at the denominator would be neglected as a result of the statement hence the equation becomes

                v = \frac{V_{max}[S]}{[S]} = V_{max}

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing [E_{total}] will lower K_M

This is because K_M does not depend on enzyme concentration it is a property of a enzyme

             

       

7 0
3 years ago
In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white soli
Digiron [165]

Explanation:

When magnesium metal burns is heated i the air it forms magnesium oxide.The balanced chemical reaction is given as:

2Mg+O_2\rightarrow 2MgO

2 moles of magnesium metal when reacts with 1 moles of oxygen it gives 2 moles of magnesium oxide which is white in color.

Some times along with formation of magnesium oxide small amount of magnesium nitride also produced due to which magnesium oxide appears grey in color .The balanced chemical reaction is given as:

3Mg+N_2\rightarrow Mg_3N_2

3 moles of magnesium combines with 1 mol of nitrogen gas to to give 1 mol of magnesium nitride.

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Convert 3841 cL to L. SHOW WORK PLS
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Answer:

1 cl = 1 * 10-5 m³

1 L = 1 * 10-3 m³

1 L = 0.001 m³

1 m³ = (1/0.001) L

1 m³ = 1000 L

1 cl = 1 * 10-5 * 1000 L

1 cl = 0.01 L

4 0
2 years ago
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