1mol ------------ 3×6,02×10²³ oxygen atoms
x ------------ 2,55×10²⁴
x = 2,5×10²⁴ × 1mol / 3×6,02×10²³ ≈ 0,138*10 mol = 1,38mol
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Answer: Option (C) is the correct answer.
Explanation:
The elements of group 2 are also known as alkaline earth metals. Thus, each element of group 2 is an alkaline earth metal.
Group 2 contains six elements from to bottom which are as follows.
Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba) and Radium (Ra). Out of which radium is a radioactive element.
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
Think of what's your vocabulary is about and try to make it into 3D