Answer:
E = 3.77×10⁻¹⁹ J
Explanation:
Given data:
Wavelength of absorption line = 527 nm (527×10⁻⁹m)
Energy of absorption line = ?
Solution:
Formula:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = speed of wave = 3×10⁸ m/s
by putting values,
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m
E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m
E = 0.0377×10⁻¹⁷ J
E = 3.77×10⁻¹⁹ J
<span>The
kingdom, protista’s characteristics are that the organism (not a plant,
animal or fungus) are:
unicellular however some are multicellular like algae, are heterotrophic or
autotrophic, others lives in water while some live in moist areas or human body,
have a nucleus, cellular respiration is primarily aerobic, some are pathogenic
(e.g. causing Malaria) and reproduction is mitosis or meiosis. This kingdom
includes: Sacordinians – pseudopods (e.g. Amoeba, Foraminiferans<span>.)</span>, Zooflagellates – flagellates
(e.g. Trypanosoma gambiense),
Ciliaphorans – ciliates (e.g. paramecium) and Sporozoans (e.g. Plasmodium).</span>
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
The answer is A) KF. The rest are covalent.
