The first-order decay of radon has a half-life of 3.823 days. how many grams of radon decomposes after 5.55 days if the sample initially weighs 100.0 grams? 83.4 g50.0 g36.6 g63.4 g16.6 g

The mass of radon that decompose = 63. 4 g

Explanation:

R.R = P.E/(2ᵇ/ⁿ)

Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.

Where P.E = 100 g , b = 5.55 days, n = 3.823 days.

∴ R.R = 100/

  R.R = 100/

  R.R = 100/2.73

  R.R = 36.63 g.

The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.

Mass of radon that decompose = 100 - 36.63

= 63.37 ≈ 63.4 g

The mass of radon that decompose = 63. 4 g

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm