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choli [55]
3 years ago
7

What is the entropy change associated with the expansion of one mole of an ideal gas from an initial volume of v to a final volu

me of v of 2.50v at constant temperature?
Physics
1 answer:
Kaylis [27]3 years ago
4 0

Answer:

ΔS = 7.618 J/K

Explanation:

The entropy change associated to an isothermal process is:

ΔS = \frac{Q}{T}

We can calculate Q (heat) how:

Q = nRTln(\frac{Vf}{Vi})

   Where Vf = 2.5v

               Vi = v

               R = 8.314J/mol.k  (Constant of ideal gases)

               n: Number of moles = 1 mol

Then

Q = nRTln(2.5)

ΔS = \frac{nRTln(2.5)}{T}

ΔS = (1)(8.314)ln(2.5)

ΔS = 7.618 J/K

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What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL?<br> Show work!
Luden [163]

Answer:

The answer is 21024g/mL

Explanation:

Multiply 657 by 32:

 657

×  32

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

   ↳            21024 g/mL

3 0
3 years ago
An atom has 5 protons, 6 neutrons, and 5 electrons. what is the atomic mass?​
ozzi

Answer:

the atomic mass is 11

Explanation:

the atomic mass is basically how many protons and neutrons there are so for this all you have to do is some simple math:

5 + 6 = 11

and boom, ur atomic mass is equal to 11!

4 0
2 years ago
Which is most likely the length of a student’s textbook?
UkoKoshka [18]

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

30 mm = (30) * (\frac{1}{1000}) = 0.030 m

30 cm = (30) * (\frac{1}{100}) = 0.30 m

30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m

Therefore, physically the correct measure is:

0.30 m = 30 cm

Answer:

the length of a student's textbook most likely is:

30 centimeters

3 0
3 years ago
Read 2 more answers
An amateur astronomer grinds a double convex lens whose surfaces have radii of curvature of 40 cm and 60 cm. The glass has an in
DedPeter [7]

Answer:

44.4cm

Explanation:

glass has an index of refraction .n = 1.54

radii of curvature of 40 cm R1 = 40 by

radii of curvature of 600 cm R2 = 60

Now, by lens maker formula

1/f = (n - 1) (1/R1 - 1/R2)

Putting in the given values for n = 1.54 , we get f = 22.2

\frac{1}{f} = (1.54 -1) (\frac{1}{40cm} -\frac{1}{(-60cm)} )

\frac{1}{f} = 0.0225

f = 1 / 0.0225

f = 44.4cm

so, focal length in air will be  = 44.4 cm

6 0
3 years ago
A concert loudspeaker suspended high off the ground emits 31 W of sound power. A small microphone with a 1.0 cm^2 area is 42 m f
Aleonysh [2.5K]

Answer:

intensity of sound at level of microphone is 0.00139 W / m 2

sound intensity level at position of micro phone is 91.456 dB

EXPLANATION:

Given data:

power of sound   P = 31 W

distance betwen microphone & speaker is   42 m

a) intensity of sound at microphone is calculated as

                   I = \frac{P}{A}

                   = \frac{34}{4 \pi ( 44m )^ 2}

                   =  0.00139 W / m 2        

b) sound intensity level at position of micro phone is

                \beta = 10 log \frac{I}{I_o}

    where I_o id reference sound intensity and taken as

                = 1 * 10^{-12} W / m 2  

               \beta = 10 log\frac{0.00139}{10^[-12}}

                     = 91.456 dB

7 0
3 years ago
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