Respuesta :
Answer: 1.1347 * 10 ^ - 8 liter / day
Explanation:
1) convert 9.64 g of Ra to number of moles, using the atomic mass of Ra
number of moles = mass in grams / atomic mass
atomic mass of Ra = 226 g/mol
number of moles Ra = 9.64 g / 226 g/mol = 0.042654867 moles
2) number of atoms in 0.042654867 moles
number of atoms = number of moles * Avogadro's number
number of atoms of Ra = 0.042654867 * 6.022 * 10^ 23 = 2.568676 * 10^ 22 atoms of Ra
3) atoms of Rn produced
Proportion
1.00 × 10^15 atoms of Ra 2.568676 × 10^22 atoms of Ra
------------------------------------- = --------------------------------------------
1.373 × 10^4 atoms of Rn/s x
x = 3.52679 × 10^ 10 atoms of Rn/s
4) liters per day
PV = nRT => V = nRT /P
STP => T = 273.15K, P = 1 atm
n = 3.52679 × 10^ 10 atoms of Rn/s / (6.022 * 10^ 23 atoms / mol) =5.85651 * 10 ^ -14 mol/s
=> V = 5.85651 * 10^ -14 mol/s * 0.0821 atm*liter / mol*K * 273.15 / 1 atm =
V = 1.3133 * 10 ^ -13 liter / s
Per day => 1.313 * 10 ^ -13 liter / s * 24 h/day * 3600 s/h = 1.1347 * 10 ^ - 8 liter / day
And that is the answer:
Explanation:
1) convert 9.64 g of Ra to number of moles, using the atomic mass of Ra
number of moles = mass in grams / atomic mass
atomic mass of Ra = 226 g/mol
number of moles Ra = 9.64 g / 226 g/mol = 0.042654867 moles
2) number of atoms in 0.042654867 moles
number of atoms = number of moles * Avogadro's number
number of atoms of Ra = 0.042654867 * 6.022 * 10^ 23 = 2.568676 * 10^ 22 atoms of Ra
3) atoms of Rn produced
Proportion
1.00 × 10^15 atoms of Ra 2.568676 × 10^22 atoms of Ra
------------------------------------- = --------------------------------------------
1.373 × 10^4 atoms of Rn/s x
x = 3.52679 × 10^ 10 atoms of Rn/s
4) liters per day
PV = nRT => V = nRT /P
STP => T = 273.15K, P = 1 atm
n = 3.52679 × 10^ 10 atoms of Rn/s / (6.022 * 10^ 23 atoms / mol) =5.85651 * 10 ^ -14 mol/s
=> V = 5.85651 * 10^ -14 mol/s * 0.0821 atm*liter / mol*K * 273.15 / 1 atm =
V = 1.3133 * 10 ^ -13 liter / s
Per day => 1.313 * 10 ^ -13 liter / s * 24 h/day * 3600 s/h = 1.1347 * 10 ^ - 8 liter / day
And that is the answer:
The amount of Rn produced per day by 9.64 g of Ra is 1.13 [tex]\rm \times\;10^-^8[/tex] L.
Moles of Radon (Ra) in 9.64 g of Ra are;
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of Ra = [tex]\rm \dfrac{9.64}{226}[/tex]
Moles of Ra = 0.0426
Number of atoms of Ra in 0.0426 moles of Ra are:
Number of atoms = moles [tex]\times[/tex] Avagadro number
= 0.426 [tex]\times\;6.203\;\times\;10^2^3[/tex]
number of atoms of Ra produced by 9.64 g of Ra are 2.56 [tex]\rm \times\;10^2^3[/tex]
the proportion of atoms : atom\sec will be:
[tex]\rm \dfrac{atoms\;of\;Ra}{atoms\;of\;Ra/sec}\;=\;\dfrac{atoms\;of\;Rn}{x}[/tex]
[tex]\rm \dfrac{1\;\times\;10^1^5}{1.373\;\times\;10^4}\;=\;\dfrac{2.56\;\times\;10^2^2}{x}[/tex]
Atoms of Rn produced per second are 3.53 [tex]\rn \times\;10^1^0[/tex] atoms/sec.
Moles of Rn produced = [tex]\rm \dfrac{atoms\;per\;sec}{avagadro\;number}[/tex]
Moles of Rn produced = [tex]\rm \dfrac{3.52\;\times\;10^1^0}{6.023\;\times\;10^2^3}[/tex] moles
Moles of Rn produced = 5.85 [tex]\rm \times\;10^-^1^4[/tex] mol/sec.
From the ideal gas equation,
PV = nRT
The volume of Rn produced = [tex]\rm \dfrac{nRT}{P}[/tex]
= [tex]\rm \dfrac{5.85\;\times\;10^-^1^4\;\times\;0.0821\;\times\;273.15}{1}[/tex]
= 1.31 [tex]\rm \times\;10^-^1^3[/tex] liter/sec.
The amount of Rn produced per day = amount produced per second [tex]\times[/tex] 3600
The amount of Rn produced per day = 1.31 [tex]\rm \times\;10^-^1^3[/tex] [tex]\times[/tex] 24 [tex]\times[/tex] 3600 L
The amount of Rn produced per day by 9.64 g of Ra is 1.13 [tex]\rm \times\;10^-^8[/tex] L.
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