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>
We can't answer without the following
If we assume the density of water= 1000 gm/lit.
so Volume = Mass / Density
Vol. = 524.24/1000
Vloume = 0.52424 liters
The balanced equation for the reaction is as follows
2Na + Cl₂ --> 2NaCl
first we need to find the limiting reactant from the 2 reactants in the reaction
stoichiometry of Na to Cl₂ is 2:1
number of Na moles - 133.0 g / 23 g/mol = 5.783 mol
at STP, molar volume is where 1 mol of any gas occupies a volume of 22.4 L
if 22.4 L occupied by 1 mol
then 33.0 L is occupied by 33.0 L / 22.4 L/mol = 1.47 mol
therefore number of Cl₂ moles present - 1.47 mol
If Cl₂ is the limiting reactant
1 mol of Cl₂ reacts with 2 mol of NaCl
therefore 1.47 mol of Cl₂ reacts with - 2 x 1.47 = 2.94 mol of Na
but 5.783 mol of Na is present which means Na is in excess and Cl₂ is the limiting reactant
NaCl formed depends on amount of Cl₂ present
stoichiometry of Cl₂ to NaCl is 1:2
number of NaCl moles formed - 2 x 1.47 mol = 2.94 mol of NaCl
mass of NaCl formed - 2.94 mol x 58.5 g/mol = 172 g
172 g of NaCl is formed