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.
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Answer:
Arts and science are similar in that they are expressions of what it is to be human in this world.
Explanation:
Both are driven by curiosity, discovery, the aspiration for knowledge of the world or oneself, and perhaps, as the conceptual artist Goshka Macuga said on her recent visit to Cern, a desire for world domination.
The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula
% yield = actual yield/ theoretical yield x100
actual yield =575 grams
to calculate theoretical yield
find the moles of SO3 used =mass/molar mass
= 500g/ 80 g/mol =6.25 moles
SO3+H2O=H2SO4
by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also= 6.25 moles
the theoretical yield of H2SO4 is therefore = moles /molar mass
= 6.25 x98= 612.5 grams
%yield is therefore= 575 g/612 g x100= 93.9 %
Answer:
The energy released will be -94.56 kJ or -94.6 kJ.
Explanation:
The molar mass of methane is 16g/mol
The given reaction is:
the enthalpy of reaction is given as ΔH = -890.0 kJ
This means that when one mole of methane undergoes combustion it gives this much of energy.
Now as given that the amount of methane combusted = 1.70g
The energy released will be:
Over the last few days, NASA has been getting everybody excited with its plans to hold a high profile briefing about a discovery beyond our solar system. Now we know that the announcement concerns the discovery of seven planets orbiting a star elsewhere in our galaxy<span>. hope this helps</span>
