<span>Calcitriol </span>is the active form of vitamin D. (D)
The half life of carbon-14 is 5700 years. (Half life is the time taken by a radioactive isotope to decay by half of its original mass).
Let A₀ be the initial amount of carbon-14 that is found in living matter (t=0 years), to determine when there was 44.5% of A₀ left.
44.5 = 100 × (1/2)^n, where n is the number of half lives
0.5^n = 0.445
n = log 0.445/log 0.5
n = 1.168
But 1 half life is 5700 years
Therefore, the number of years will be 5700 × 1.168 = 6658.299725 years
≈ 6658.30 years
Answer:
0.208mole of CO2
Explanation:
First, let us calculate the number of mole of HC3H3O2 present.
Molarity of HC3H3O2 = 0.833 mol/L
Volume = 25 mL = 25/100 = 0.25L
Mole =?
Mole = Molarity x Volume
Mole = 0.833 x 0.25
Mole of HC3H3O2 = 0.208mole
Now, we can easily find the number of mole of CO2 produce by doing the following:
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
From the equation,
1mole of HC2H3O2 produced 1 mole of CO2.
Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2
Answer:
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Answer: The box is moving downward with increasing speed.
Explanation:
