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valentina_108 [34]
3 years ago
13

A vat of nitrogen at its boiling point (-196 degrees C) absorbs 384000 J of heat. How much mass of nitrogen burns off? (Unit = k

g)
Physics
1 answer:
masya89 [10]3 years ago
3 0

Answer:

1.92 kg of nitrogen.

Explanation:

The following data were obtained from the question:

Heat absorbed (Q) = 384000 J

Note: Heat of vaporisation (ΔHv) of nitrogen = 5600 J/mol

Next, we shall determine the number of mole of nitrogen that absorbed 384000 J.

This is illustrated below:

Q = mol·ΔHv

384000 = mole of N2 x 5600

Divide both side by 5600

Mole of N2 = 384000/5600

Mole of N2 = 68.57 moles

Next, we shall convert 68.57 moles of nitrogen, N2 to grams.

This can be obtained as follow:

Molar mass of N2 = 2 x 14 = 28 g/mol.

Mole of N2 = 68.57 moles.

Mass of N2 =..?

Mole = mass /molar mass

68.57 = mass of N2 /28

Cross multiply

Mass of N2 = 68.57 x 28

Mass of N2 = 1919.96 g

Finally, we shall convert 1919.96 g to kilograms.

This can be achieved as shown below:

1000g = 1 kg

Therefore,

1919.96 g = 1919.96/1000 = 1.92 kg.

Therefore, 1.92 kg of nitrogen were burned off.

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<em> </em><em> </em><em>62.125</em>

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6 0
3 years ago
If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol
Mrrafil [7]

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation \theta=3.0\times10^{-5}\ rad

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

\sin\theta=\dfrac{1.22\lambda}{d}

\lambda=\dfrac{d\sin\theta}{1.22}

Put the value into the formula

\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}

For small angle \sin\theta\approx\theta

\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}

\lambda=4.92\times10^{-7}\ m

\lambda=492\ nm

Hence, The maximum wavelength is 492 nm.

5 0
3 years ago
I really need help please just answer at least one
yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0
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Allison is looking through a telescope at an object in space. The object looks like a very small planet, and it does not have a
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Answer:

Allison is probably looking at the asteroid.

Explanation:

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  • Asteroids orbits around the sun in very irregular ans strange path unlike the other planets.
  • This is also the reason why it's called as the minor planets.
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True or false the hotter the star the higher is absolute magnitude?
liubo4ka [24]

Answer:

true

Explanation:

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