Answer: Option (3) is the correct answer.
Explanation:
Aerobic organisms are the organisms which survive and grow in the presence of oxygen.
When oxidation of glucose occurs in the presence of oxygen then it is known as aerobic respiration.
In aerobic respiration, food releases energy to produce ATP which is necessary for cell activity. There is complete breakdown of glucose in aerobic respiration that is why more energy is released. Therefore, aerobic organisms become active.
Thus, we can conclude that characteristics very active, efficient use of energy describes aerobic organisms.
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
Yes, because CO2 is carbon dioxide, and carbon dioxide contains molecules.
Answer:
B. Ca2+ import into the ER because it has the steeper concentration gradient
Explanation:
ΔGt = RT㏑(C₂/C₁)
where ΔGt is the free energy change for transport; R = 8.315 J/mol; T = 298 K; C₂/C₁ is ratio of concentrations inside and outside each organelle.
For Ca²⁺ import
ΔGt = 8.315 J/mol * 298 K * ㏑(10⁻³/10⁻⁷)
ΔGt= 3.42 kJ/mol
For H⁺ import
ΔGt = 8.315 J/mol * 298 K * ㏑ (10⁻⁴/10⁻⁷)
ΔGt = 2.73 kJ/mol
From the above values, ΔGt is greater for Ca²⁺ import because it has a steeper concentration gradient
