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Elena L [17]
4 years ago
13

The half-life of radon-222 is 3.82 days. If a sample of gas contains 4.38 μg of radon-222, how much will remain in the ample aft

er 15.2 days?
Chemistry
1 answer:
LekaFEV [45]4 years ago
3 0

Answer:

0.28 \mu g

Explanation:

We can write the amount of mass of radon-222 left after a time t by using the equation:

m(t) = m_0 (\frac{1}{2})^{t/t_{1/2}}

where

m_0 = 4.38 \mu g is the initial mass

t is the time

t_{1/2} = 3.82 d is the half-life

Substituting t = 15.2 d in the formula, we find

m(15.2 d) = (4.38 \mu g) (\frac{1}{2})^{15.2d/3.82 d}=0.28 \mu g

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A saturated solution is made by dissolving 0.327 g of a polypeptide (a substance formed by joining together in a chainlike fashi
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Answer: The approximate molecular mass of the polypeptide is 856 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L)}}\times RT

where,

\pi = osmotic pressure of the solution = 4.19 torr

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polypeptide) = 0.327 g

Volume of solution = 1.70 L

R = Gas constant = 62.364\text{ L.torr }mol^{-1}K^{-1}

T = temperature of the solution = 26^oC=[273+26]K=299K

Putting values in above equation, we get:

4.19torr=1\times \frac{0.327}{\text{Molar mass of solute}\times 1.70}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\\text{molar mass of solute}=856g/mol

Hence, the molar mass of the polypeptide is 856 g/mol

8 0
3 years ago
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