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VikaD [51]
2 years ago
8

At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +

2 B ( g ) ⟶ 5 C ( g ) 2A(g)+2B(g)⟶5C(g) A ( g ) + B ( g ) ⟶ C ( g ) A(g)+B(g)⟶C(g) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )
Chemistry
1 answer:
Tju [1.3M]2 years ago
4 0

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

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what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c
slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

Q = m \times C \times \bigtriangleup T

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

C = \frac{Q}{m \times \bigtriangleup T}

Let us plug in the values in above equation.

C = \frac{300J}{267g \times 12 C}

C = \frac{300J}{3204 g C}

C = 0.0936 J/g °C

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3 0
3 years ago
For the chemical reaction:
Kamila [148]

Answer:

Volume of ammonia produced = 398.7 dm³

Explanation:

Given data:

Volume of N₂ = 200 dm³

Pressure and temperature = standard

Volume of ammonia produced = ?

Solution:

Chemical equation:

N₂ + 3H₂     →      2NH₃

Number of moles of N₂:

PV = nRT

1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K

n = 200 atm.L /22.41 atm.L/mol

n = 8.9 mol

Now we will compare the moles of ammonia and nitrogen.

               N₂          :         NH₃

                1            :           2

              8.9          :        2/1×8.9 = 17.8 mol

Volume of ammonia:

1 mole of any gas occupy 22.4 dm³ volume

17.8 mol ×22.4 dm³/1 mol = 398.7 dm³

7 0
3 years ago
Assuming that both cases describe hydrogen‑like atoms with one electron, for which case is more energy emitted or absorbed?
Nana76 [90]

Assuming that both cases describe hydrogen‑like atoms with one electron, More energy is emitted or absorbed for case 2. The correct option is D.

<h3>What is emitting of energy, by electron?</h3>

The energy of the electron decreases as it changes levels, and emission of photons happens in the atom.

With the electron moving from a higher to a lower energy level, the photon is emitted. The photon's energy is the same as the energy lost by an electron moving to a lower energy level.

Thus, the correct option is D, More energy is emitted or absorbed for case 2.

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5 0
1 year ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
2 years ago
The overall cell reaction occurring in an alkaline battery isZn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s) (e) In practice, vol
zubka84 [21]

b) Mass of MnO₂ = 5.981 g

    Mass of H₂O = 1.2384 g

c) Total Mass of Reactant consumed = 11.708 g

b) Given Reaction

            Zn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s)

  Mass of Zn = 4.50 g

  Moles of Zn = 0.0688 moles

   Now,

    Moles of Zn = moles of MnO₂ = moles of H₂O = moles of ZnO = moles of                   Mn(OH)₂

Hence ,

Moles of MnO₂ = 0.0688 moles

Mass of MnO₂ = 0.0688 × 86.9368 g

                        = 5.981 g

Similarly,

    Moles of H₂O = 0.0688 moles

     Mass of H₂O = 0.0688 × 18 g

                           = 1.2384 g

c) now ,

    Moles of  ZnO = 0.0688 moles

     Mass of  ZnO  = 0.06880× 81.3794 g

                              = 5.598 g

 Moles of  Mn(OH)₂ = 0.0688 moles

  Mass of  Mn(OH)₂ =0.0688 × 88.952 g

                                  = 6.11 g

Total mass of Product = 11.708 g

Total Mass of Reactant = 11.715 g

Hence,

    Total mass of reactant consumed = 11.708 g

c)  As total mass of reactant is more than that of mass of reactant consumed , Hence G is more than that of mass of reactant consumed .

G = - nFEcell

 

and no. of moles of reactant  is greater than that of number of moles of reactant consumed .

       Hence voltaic cell of given Capacity are heavier than that of mass of reactant consumed .

 Thus from above conclusion we can say that , Mass of the reactant consumed is 11.708 g.

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3 0
1 year ago
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