Answer:
v = 7.85 × 10¹⁴ Hz
Explanation:
The first three energy level can be represented as follows:
⇅ ------> n₄ = 4
⇅ ------> n₃ = 3
⇅ ------> n₂ = 0
ΔE = hv = 
ΔE = hv = 
ΔE = hv = 
ΔE = hv = 
where h = planck constant =
J.s
mass (m) = 
e = 0.904 nm = 
hv = 
hv = 
hv = 
hv =
J
v = 
v = 
v = 
v = 7.85 × 10¹⁴ Hz (since s⁻¹ is equivalent to 1 Hz)
Thus, the frequency of an electronic transition from the highest-occupied orbital to the lowest-unoccupied orbital = 7.85 × 10¹⁴ Hz
Answer:
The answer is
<h2>1.336 atm</h2>
Explanation:
In order to find the volume of the gas at 175 mL we use the Boyle's law formula
That's

where
P1 is the initial pressure
V1 is the initial volume
V2 is the final volume
P2 is the final pressure.
Since we are finding the final pressure

From the question
P1 = 0.935 atm
V1 = 250.0 mL
V2 = 175 mL
Substitute the values into the above formula and solve
That's

We have the final answer as
<h3>1.336 atm</h3>
Hope this helps you
Answer:
The answer is A hope this helps
Explanation:
Cytosine (C) and Guanine (G) are more steady under expanding heat since C and G have three hydrogen bonds while Adenine (An) and Thymine (T) have just two. The more hydrogen bonds there are, the more steady the nucleotides are. More bond dependability requires more warmth vitality to separate the securities, and since G≡C have more hydrogen securities than A=T, they are thusly more steady under expanding heat.