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Nana76 [90]
3 years ago
13

A cylinder with a moving piston expands from an initial volume of 0.350 L against an external pressure of 1.90 atm. The expansio

n does 287 J of work on the surroundings. What is the final volume of the cylinder
Chemistry
1 answer:
myrzilka [38]3 years ago
7 0

Answer:

1.84 L

Explanation:

Using the equation for reversible work:

W = -P*(V_{2} - V_{1})

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K);  R = 0.0821 (L*atm)/(mol*K)

Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J

V_{1} is the initial volume = 0.350 L

V_{2} is the final volume = ?

Thus:

(-287 J)*0.00987 (L*atm)/J = -1.9 atm*(V_{2} - 0.350) L

V_{2} = [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L

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3 years ago
SCIENCE QUESTION plz helpz
kvasek [131]

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5 0
3 years ago
The following reaction was carried out in a 3.00 L reaction vessel at 1100 K:
Nady [450]

Answer:

Q=0.840

Explanation:

Hello,

In this case, since the given undergoing chemical reaction is correctly balanced, the reaction quotient is computed as well as the equilibrium constant but in terms of the given concentrations that are:

C_{C}=\frac{5.25mol}{3.00L} =1.75M\\C_{H_2O}=\frac{12.2mol}{3.00L} =4.07M\\C_{CO}=\frac{3.90mol}{3.00L}=1.30M\\C_{H_2}=\frac{7.90mol}{3.00L} =2.63M

In such a way, the reaction quotient turns out:

Q=\frac{C_{CO}C_{H_2}}{C_{H_2O}}=\frac{1.30M*2.63M}{4.07M}\\ Q=0.840

Taking into account that carbon is not included since it is solid.

Best regards.

7 0
2 years ago
A 45.0-gram sample of copper metal was heated from 20.0°C to 100.0°C. Calculate the heat absorbed, in kJ, by the metal.
s2008m [1.1K]

Answer:

1.386 KJ

Explanation:

From the question given above, the following data were obtained:

Mass (M) of copper = 45 g

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Heat absorbed (Q) =..?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

Change in temperature (ΔT) =?

ΔT = T2 – T1

ΔT = 100 – 20

ΔT = 80 °C

Next, we shall determine the heat absorbed by the sample of copper as follow:

Mass (M) of copper = 45 g

Change in temperature (ΔT) = 80 °C

Specific heat capacity (C) of copper = 0.385 J/gºC

Heat absorbed (Q) =..?

Q = MCΔT

Q = 45 × 0.385 × 80

Q = 1386 J

Finally, we shall convert 1386 J to KJ. This can be obtained as follow:

1000 J = 1 KJ

Therefore,

1386 J = 1386 J × 1 KJ /1000 J

1386 J = 1.386 KJ

Thus, the heat absorbed by the sample of the sample of copper is 1.386 KJ.

5 0
2 years ago
How does the size of ions affect the conductivity of a solution?
mrs_skeptik [129]

Small ions have small areas. There is less resistance as they move through the solution.

For example, in molten salts, the conductivity of <span>Li+</span> is greater than that of <span>Cs+</span>.

Small ions have high charge density.

4 0
3 years ago
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