A tsunami.Because tsunamis are unpredictable in a way, a body of water that is very vast can cause an uprising of water. As for instance, if I lived in Hawaii, and their was a tsunami coming forth from each side, it would be a problem.
Ionic Crystal. Ionic hydrogen.postive electron transfer to atoms
Answer:
46.9mg of oxygen
Explanation:
From Henry's law,
Concentration of oxygen (C) = Henry's constant (K) × partial pressure of oxygen in air (p)
K = 1.3×10^-3M/atm O2, p = mole fraction of oxygen in air × pressure of air = 0.21×1.08atm = 0.2268atm
C = K×p = 1.3×10^-3 × 0.2268 = 0.00029484M of O2
Concentration (C) = number of moles of oxygen (n)/volume of water (V)
Volume of water (V) = 4.97L
n = CV = 0.00029484 × 4.97 = 0.001465mole
number of moles (n) = mass of O2/MW of O2
mass of O2 = number of moles of O2 × MW of O2 = 0.001465mole × 32g/mole = 0.0469g = 0.0469×1000mg = 46.9mg (to three significant figures)
there is no answer to your question because all of the options ARE blood types so there is no telling.
Answer:
Percentage of a sample remains after 60.0 min is 13.03%.
Explanation:
- It is known that the decay of isotopes of C-11 obeys first order kinetics.
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- Half-life time (t1/2) in first order reaction = 0.693/k, where k is the rate constant.
∴ k = 0.693/(t1/2) = 0.693/(20.4 min) = 0.03397 min⁻¹.
- The integrated law for first order reaction is:
<em>kt = ln[A₀]/[A],</em>
where, k is the rate constant (k = 0.03397 min⁻¹).
t is the time of the reaction (t = 60.0 min).
[A₀] is the initial concentration of C-11 ([A₀] = 100.0 %).
[A] is the remaining concentration of C-11 ([A] = ???%).
<em>∵ kt = ln[A₀]/[A]</em>
∴ (0.03397 min⁻¹)(60.0 min) = ln(100%)/[A]
∴ 2.038 = ln(100%)/[A]
∴ 7.677 = (100%)/[A]
<em>∴ [A] </em>= (100%)/(7.677) = <em>13.03%.</em>
<em>So, percentage of a sample remains after 60.0 min is 13.03%.</em>