i need help solving this. Can you guys show me how it's done.
Radium-226 decays by alpha emission to radon-222, a noble gas.What volume of pure Radon- 222 at 23oC and 785mmHg could be obtained from 543.0 mg of radium bromide,RaBr2 in a period of 20.2 years.The half-life of radium-226 is 1602 years.
k = 0.693/t1/2
plug in the half life of Ra-226 and calculate k. THEN, convert 543.0 mg RaBr2 to mg Ra. I will call that value No. Use No in the following formula to calculate N (t is the time of 20.2 years.)
ln(No/N) = kt
From N, which is mg Ra remaining, calculate how much has decayed, convert the amount decayed into mass of Rn-222 and use PV = nRT to calculate the volume Rn at the conditions specified. Check my thinking.
To solve this problem, we can follow these steps:
Step 1: Calculate the decay constant (k) using the formula k = 0.693 / t_1/2, where t_1/2 is the half-life of Ra-226. Given that the half-life of Ra-226 is 1602 years, we can calculate k as follows:
k = 0.693 / 1602 = 0.0004324 (approx.)
Step 2: Convert the mass of RaBr2 (543.0 mg) to the initial mass of Ra (No) by dividing it by the molar mass of RaBr2. To do this, you need to know the molar mass of RaBr2. Let's assume it is 310.0 g/mol. Therefore:
No = 543.0 mg / 310.0 g/mol = 1.75 mg
Step 3: Apply the radioactive decay formula to calculate the remaining mass of Ra (N) after a time of 20.2 years using the formula ln(No / N) = kt. Rearranging the formula, we get:
N = No * e^(-kt)
Substituting the values, we have:
N = 1.75 mg * e^(-0.0004324 * 20.2 years)
Step 4: Calculate the mass of Rn-222 produced by subtracting N from No, which represents the decayed Ra:
Decayed Ra mass = No - N
Step 5: Convert the mass of decayed Ra to moles of Rn-222 by dividing it by the molar mass of Rn-222. Assume the molar mass of Rn-222 is 222.0 g/mol:
Moles of Rn-222 = Decayed Ra mass / 222.0 g/mol
Step 6: Use the ideal gas law, PV = nRT, to calculate the volume of Rn-222 at the given conditions. Given that the temperature (T) is 23°C (which is 296 K) and the pressure (P) is 785 mmHg, we can rearrange the formula as follows:
V = (nRT) / P
Substituting the values, we get:
V = (Moles of Rn-222 * 0.0821 L·atm/(mol·K) * 296 K) / 785 mmHg
Step 7: Calculate the volume of Rn-222 in liters by converting from millimeters of mercury (mmHg) to atmospheres (atm). Use the conversion factor 1 atm = 760 mmHg:
V_liters = V / (760 mmHg / 1 atm)
That's it! By following these steps, you should be able to calculate the volume of pure Rn-222 obtained from the given amount of RaBr2 after the specified period of time.