Chemistry, 29.07.2019 13:30 aurelio1121

Radon (rn) is the heaviest and the only radioactive member of group 8a(18), the noble gases. it is a product of the disintegration of heavier radioactive nuclei found in minute concentrations in many common rocks used for building and construction. in recent years, health concerns about the cancers caused from inhaled residential radon have grown. if 1.00 × 1015 atoms of radium (ra) produce an average of 1.373 × 104 atoms of rn per second, how many liters of rn, measured at stp, are produced per day by 9.64 g of ra?

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Answer from: dean91

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 from: Katelynn7086

The amount of Rn produced per day by 9.64 g of Ra is 1.13 $\rm \times\;10^-^8$ L.

Moles of Radon (Ra) in 9.64 g of Ra are;

Moles = $\rm \dfrac{weight}{molecular\;weight}$

Moles of Ra = $\rm \dfrac{9.64}{226}$

Moles of Ra = 0.0426

Number of atoms of Ra in 0.0426 moles of Ra are:

Number of atoms = moles $\times$ Avagadro number

= 0.426 $\times\;6.203\;\times\;10^2^3$

number of atoms of Ra produced by 9.64 g of Ra are 2.56 $\rm \times\;10^2^3$

the proportion of atoms : atom\sec will be:

$\rm \dfrac{atoms\;of\;Ra}{atoms\;of\;Ra/sec}\;=\;\dfrac{atoms\;of\;Rn}{x}$

$\rm \dfrac{1\;\times\;10^1^5}{1.373\;\times\;10^4}\;=\;\dfrac{2.56\;\times\;10^2^2}{x}$

Atoms of Rn produced per second are 3.53 $\rn \times\;10^1^0$ atoms/sec.

Moles of Rn produced = $\rm \dfrac{atoms\;per\;sec}{avagadro\;number}$

Moles of Rn produced = $\rm \dfrac{3.52\;\times\;10^1^0}{6.023\;\times\;10^2^3}$ moles

Moles of Rn produced = 5.85 $\rm \times\;10^-^1^4$ mol/sec.

From the ideal gas equation,

PV = nRT

The volume of Rn produced = $\rm \dfrac{nRT}{P}$

= $\rm \dfrac{5.85\;\times\;10^-^1^4\;\times\;0.0821\;\times\;273.15}{1}$

= 1.31 $\rm \times\;10^-^1^3$ liter/sec.

The amount of Rn produced per day = amount produced per second $\times$ 3600

The amount of Rn produced per day = 1.31 $\rm \times\;10^-^1^3$ $\times$ 24 $\times$ 3600 L

The amount of Rn produced per day by 9.64 g of Ra is 1.13 $\rm \times\;10^-^8$ L.

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Answer from: Quest

It d from what i know

Answer from: Quest

Icant see the pictures

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Radon (rn) is the heaviest and the only radioactive member of group 8a(18), the noble gases. it is a...

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