Answer:
As you noticed, the percent-error formula needs you to use a value in the ... The actual absolute zero is 275C below my coldest recorded value. ... The only way your calculations makes sense is you're trying to calculate how ... Next, I'll interpret "I found that AZ is at -258°C" as "I calculated AZ to be 260 degrees below 2°C".
Explanation:
9.01 × 10⁻²⁶ J
<h3>Explanation</h3>
ΔE = h · f
Where
- ΔE the change in energy,
- h the planck's constant, and
- f frequency of the emission.
However, only λ is given.
f = c / λ
Where
- f frequency of the emission,
- λ wavelength of the emission, and
- c the speed of light.
For this emission:
f = 2.998 × 10⁸ / 2.21 = 1.36 × 10⁸ s⁻¹.
ΔE = h · f = 6.626 × 10⁻³⁴ × 1.36 × 10⁸ = 9.01 × 10⁻²⁶ J
Answer:
Cl^- <S^2-<Sc^3+ <Ca^2+<K^+
Explanation:
We know that ionic radius of ions decreases from right to left in the periodic table. This is because, ionic radii decreases with increase in nuclear charge. This explains why; Sc^3+ <Ca^2+<K^+.
Secondly, even though Cl^- is isoelectronic with S^2-, the size of the nuclear charge in Cl^- is larger compared to that of S^2- . Hence Cl^- is smaller than S^2- in ionic radius owing to increased nuclear attraction in Cl^-.
Its a chemical element with symble
The half-life of polonium-210, given that it decays from 98.3 micrograms to 12.3 micrograms in 414 days is 138 days
<h3>How to determine the number of half-lives </h3>
- Original amount (N₀) = 98.3 micrograms
- Amount remaining (N) = 12.3 micrograms
- Number of half-lives (n) =?
2ⁿ = N₀ / N
2ⁿ = 98.3 / 12.3
2ⁿ = 8
2ⁿ = 2³
n = 3
<h3>How to determine the half life </h3>
- Number of half-lives (n) = 3
- Time (t) = 414 days
- Half-life (t½) = ?
t½ = t / n
t½ = 414 / 3
t½ = 138 days
Learn more about half life:
brainly.com/question/26374513
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