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3 years ago

What is the specific heat of a 44 g piece of metal if 202 Joules of heat are required to

Chemistry

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

≈ 395,8 J/(kg * °C)

Explanation:

m = 44 g = 0,044 kg

$t_{1}$ = 22 °C

$t_{2}$ = 33,6 °C

Q = 202 J

The formula is: Q = c * m * ( $t_{2} - t_{1}$ )

c = $\frac{Q}{m * (t_{2} - t_{1} )}$

Calculating:

c = 202 J / 0,044 kg * (33,6 °C - 22 °C) ≈ 395,8 J/(kg * °C)

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Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit

djverab [1.8K]

Answer:

1.64x10⁻¹⁸ J

Explanation:

By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.

When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:

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Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.

The wavelength can be calculated by:

1/λ = R*(1/nf² - 1/ni²)

Where R is the Rydberg constant (1.097x10⁷ m⁻¹), nf is the final orbit, and ni the initial orbit. So:

1/λ = 1.097x10⁷ *(1/1² - 1/2²)

1/λ = 8.227x10⁶

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4 years ago

A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam

Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

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3 years ago