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ladessa [460]
2 years ago
13

The first excited state of Ca is reached by absorption of 422.7 nm light. Find the energy difference (kJ/mole) between the groun

d and first excited state. Watch your units. A. 2.83 x 10-7 B. 4.70 x 10-22 C. 283 D. 4.70 x 10-19 E. 457
Chemistry
1 answer:
dedylja [7]2 years ago
5 0

For the excited state of Ca at the absorption of 422.7 nm light,the energy difference  is mathematically given as

E= 4.70x10-22 kJ/mol

<h3>What is the energy difference (kJ/mole) between the ground and the first excited state?</h3>

Generally, the equation for the Energy  is mathematically given as

E = nhc / λ

Where

h= plank's constant

h= 6.625x 10-34 Js

c = speed of light

c= 3x 108 m/s

Therefore

E = 1*(6.625x 10-34 Js)( 3x 10^8 m/s) / ( 422.7x10^-9)

E= 4.70x10-22 kJ/mol

In conclusion, Energy  

E= 4.70x10-22 kJ/mol

Read more about Energy

brainly.com/question/13439286

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Answer:

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Explanation:

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Answer:

<u>One lone-Pair is present in Ammonia</u>

<u></u>

Explanation:

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So, there is only 1 lone pair in the ammonia molecule .

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