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
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brainly.com/question/13439286
Lower than 7 is acid greater than 7 is a base
Answer:
1.71x10²⁷
Explanation:
If we sum 1/2 of (3) + 1/2 of (1):
1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷ = 4.58x10²³
1/2 (1) 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8
C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>
K' = 4.58x10²³ * 11.8 = 5.42x10²⁴
+1/2 (2):
<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2
C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)
K'' = 5.42x10²⁴* 316.2 =
<h3>1.71x10²⁷</h3>
Answer:
D
Explanation:
Its the only answer that actually makes sense and i got it right on the quiz.