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Bezzdna [24]
3 years ago
11

Se informa que un arco eléctrico estabilizado con agua llegó la temperatura de 25600 °F. En la escala absoluta, ¿Cuál es la rela

ción entre la temperatura y la de una llama de oxiacetileno (3500 °C)?
Chemistry
1 answer:
nignag [31]3 years ago
7 0

Answer:

Explanation:

Translation:

It is reported that an electric arc stabilized with water reached the temperature of 25600 ° F. On the absolute scale, what is the relationship between the temperature and that of an oxyacetylene flame (3500 ° C)?

Solution:

Temperature of the electric arc = 25600°F

Temperature of the oxyacetylene flame = 3500°C

The absolute temperatures are usually recorded in Kelvin. In order to compare the two given temperatures, a need to convert to Kelvin arises.

From celcius to kelvin we have:

     K = 273.15 + T°C

T°C is the temperature in degree celcius

For the oxyacetylene flame:

     K = 273.15 + 3500 = 3773.15K

From fahrenheit to kelvin, we have:

    K = \frac{5}{9}(T °F – 32) + 273.15

T °F is the temperature in degree fahrenheit

For the electric arc:

    K = \frac{5}{9}(25600 – 32) + 273.15 = 14477.59K

We can see that by expressing the two temperature on the same absolute scale, an electric arc is by far hotter than an oxyacetylene flame.

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Find the volume of a gold ring that has a mass of 12.00 grams. The density of gold is 19.30 g/mL. The volume is
kozerog [31]

Answer:

<h3>The answer is 0.622 mL</h3>

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3 0
3 years ago
calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?
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<u>Answer:</u>

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}

\Delta E=-1312[0.111-1]KJ mol^{-1}

\Delta E=1166 KJ mol^{-1}

\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}

=193623 \times 10^{-23}  \frac {J}{photon}

\Delta E=1.93623 \times 10^{-18}  \frac {J}{photon}

\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}

h is planck's constant  

c is the speed of light

λ is the wavelength of light  

\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18}  J/photon)}

Wavelength

\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m}  =103 nm (Answer)

<em>Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is </em><u><em>103 nm.</em></u>

7 0
3 years ago
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