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Maru [420]
3 years ago
14

What is the energy of a microwave photon that has a frequency of 9.86 x 1012 Hz?

Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
5 0

Answer:

6.533 × 10^-21J

Explanation:

The energy of the microwave photon can be calculated using:

E = hf

Where;

E = energy of photon (J)

h = Planck's constant (6.626 × 10^-34 J/s)

f = frequency (9.86 x 10^12 Hz)

Hence, E = hf

E = 6.626 × 10^-34 × 9.86 x 10^12

E = 65.33 × 10^(-34 + 12)

E = 65.33 × 10^(-22)

E = 6.533 × 10^-21J

The energy of the microwave photon is

6.533 × 10^-21J

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Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

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(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ

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(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.5kJ)=-787kJ

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.8kJ)=-571.6kJ

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)

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Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

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