<span>i get 3.19x10^20 atoms
</span>
<span>measurement in Ci/Bq
the amount of radioactive materials released into the environment.
number of disintegrations of radioactive atoms in a radioactive material over a period of time</span>
Refer to the attachment for answer.
SOME EXTRA INFORMATION:
Bromo is used for Bromine (Br)
Chloro is used for Chlorine(Cl)
'ene' represents double bond between carbon atoms.
'ol' is used for alcohol
HOPE IT IS USEFUL
<em>Answer :</em> 72.05 g/mol
<span>
<em>Explanation : </em>
Let's </span>assume that the given gas is an ideal gas. Then we can use ideal gas equation,<span>
PV = nRT<span>
</span>
Where,
P = Pressure of the gas (Pa)
V = volume of the gas (m³)
n = number of moles (mol)
R = Universal gas constant (8.314 J mol</span>⁻¹ K⁻¹)<span>
T = temperature in Kelvin (K)
<span>
The given data for the gas </span></span>is,<span>
P = 777 torr = 103591 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³<span>
T = (</span>126 + 273<span>) = 399 K
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
n = ?
By applying the formula,
103591 Pa x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 399 K<span>
n = 3.90 x 10</span>⁻³<span> mol
</span>Moles (mol) = mass (g) /
molar mass (g/mol)<span>
Mass of the gas = </span><span>0.281 g
</span>Moles of the gas = 3.90 x 10⁻³ mol
<span>Hence,
molar mass of the gas = mass / moles
= 0.281 g / </span>3.90 x 10⁻³ mol
<span> = 72.05 g/mol
</span>
Molar mass Li2CO3 = 73.89 g/mol
Molar mass Li = 6.94g/mol Li = 6.94*2 = 13.88g
% LI = 13.88/73.89*100 = 18.78% perfectly correct.
