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IRISSAK [1]
3 years ago
8

The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x

10 ⁻³⁴ J * 8 )
SHOW WORK
Chemistry
1 answer:
Andrews [41]3 years ago
6 0
Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

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Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of
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Answer : The amount of heat released, 45.89 KJ

Solution :

Process involved in the calculation of heat released :

(1):H_2O(l)(314K)\rightarrow H_2O(l)(273K)\\\\(2):H_2O(l)(273K)\rightarrow H_2O(s)(273K)\\\\(3):H_2O(s)(273K)\rightarrow H_2O(s)(263K)

Now we have to calculate the amount of heat released.

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = amount of heat released = ?

m = mass of water = 27 g

c_{p,l} = specific heat of liquid water = 4.184 J/gk

c_{p,s} = specific heat of solid water = 2.093 J/gk

\Delta H_{fusion} = enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole

conversion : 0^oC=273k

Now put all the given values in the above expression, we get

Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]

Q=45896.798J=45.89KJ     (1 KJ = 1000 J)

Therefore, the amount of heat released, 45.89 KJ

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3 years ago
Read 2 more answers
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2 years ago
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Answer:

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Explanation:

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4 0
3 years ago
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