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>
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Answer:
The metal which reduces the other compound is the one higher in reactivity. So in this case, it is.
Explanation:
Answer:
The solution is shown in the attached image to this answer.
Explanation:
The first attached image is the complete question.
The second attached image is the solution to the question.
Answer:
.15 grams of MgO / 40.3 = .0285 moles of MgO. For every mole of Magnesium Oxide, you have one mole of magnesium and one mole of oxygen, therefore,.0285 moles of MgO * (1 mole of magnesium/1 mole of magnesium oxide) = .0285 magnesium.0285 magnesium * (6.022 *10^23 atoms magnesium/1 mole of magnesium) = 1.718*10^22 atoms of magnesium
Explanation: