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1 year ago

Calculate the number of moles of gas present in the sealed container at a

Physics

1 answer:

aliina [53]1 year ago

8 0

Here volume of gas is not given so question is solved assuming volume as 1 L.

The number of moles of 1 L gas present in the sealed container at a

pressure of 125 kPa at 25 degrees Celsius is 0.0067 moles.

The ideal gas law equation can be written as

PV = nR T

Here

P is the pressure of the gas in atm

V is the volume it occupies in L

n is the number of moles of gas present in the sample

R is the universal gas constant, equal to 0.0821 atm L/ mol K

T is the absolute temperature of the gas in Kelvin

Now, it's important to realize that the units you have for the volume, pressure, and temperature of the gas must match the unit used in the expression of the universal gas constant.

P = 125 kPa

1 atm = 760 kPa

P = 125/760 = 0.1644 atm

T = 25 degree celsius = 25 +273 = 298 K

Taking V = 1 L

n = PV/RT

n = 0.1644 x 1 / 0.0821 x 298

n = 0.0067 moles

To learn more about the ideal gas law, please click on the link brainly.com/question/128737528

#SPJ9

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Answer:

A) Force

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It is an example of force since force is a vector quantity so it has magnitude and direction. In this case the magnitude is equal to 5 [N] and the direction is upward.

The weight can not be, as it always acts downward.

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3 years ago

Select the correct answer.

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Answer:

B. It is directly proportional to the source charge.

Explanation:

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This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

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Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

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2 years ago

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C. displacement

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3 years ago

Read 2 more answers

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

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Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I E I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I E I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

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The smaller body will have greater temperature change.

<h3>Explanation:</h3>

Temperature is defined as the degree of hotness or coldness of a body. The relationship of the temperature with heat is described as

Q =m c dT.

Where Q is the heat content

m is the mass of body

c is the specific heat of body

dT is the temperature change of body.

Here the bodies are made up of same substance, so specific heat is same. The mass of bigger body is M and smaller body is m.

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So M dT1 = mdT2.

So, M/m = dT2 / dT1.

So the the smaller body will be suffering higher temperature change.

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