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>
2.26%, 26.8%, 2 2/5, 2.62, 271%
Hope this helps! :D
Since there's specific heat, you should use Q=mc△T. Depends on if this question also involves phase change or not, you might will need Lf (latent heat of fusion) or Lv (latent heat of vaporisation).
Answer:
Explanation:
Molal freezing point depression constant of butanol Kf = 8.37⁰C /m
ΔTf = Kf x m , m is no of moles of solute per kg of solvent .
mol weight of butanol = 70 g
235.1 g of butanol = 235.1 / 70 = 3.3585 moles
3.3585 moles of butanol dissolved in 4.14 kg of water .
ΔTf = 8.37 x 3.3585 / 4.14
= 6.79⁰C
Depression in freezing point = 6.79
freezing point of solution = - 6.79⁰C .
PV / T = P'V' / T'
V = V'
P / T = P' / T'
P = 630 mmHg
T = 100 K
P' = 1760 mmHg
T' = ?
630 / 100 = 1760 / T'
T' = 1760 / 6,3
T' = 279,36 K
T' ≈ 280 K
