Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
Answer:
The molecular formula of estradiol is:
.
Explanation:
Molar mass of of estradiol = M= 272.37 g/mol
Let the molecular formula of estradiol be ![C_xH_yO_z](https://tex.z-dn.net/?f=C_xH_yO_z)
Percentage of an element in a compound:
![\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BNumber%20of%20atoms%20of%20element%7D%5Ctimes%20%5Ctext%7BAtomic%20mass%20of%20element%7D%7D%7B%5Ctext%7Bmolecular%20mass%20of%20element%7D%7D%5Ctimes%20100)
Percentage of carbon in estradiol :
![\frac{x\times 12 g/mol}{272.37 g/mol}\times 100=79.37\%](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5Ctimes%2012%20g%2Fmol%7D%7B272.37%20g%2Fmol%7D%5Ctimes%20100%3D79.37%5C%25)
x = 18.0
Percentage of hydrogen in estradiol :
![\frac{y\times 1g/mol}{272.37 g/mol}\times 100=8.88\%](https://tex.z-dn.net/?f=%5Cfrac%7By%5Ctimes%201g%2Fmol%7D%7B272.37%20g%2Fmol%7D%5Ctimes%20100%3D8.88%5C%25)
y = 24.2 ≈ 24
Percentage of oxygen in estradiol :
![\frac{z\times 16 g/mol}{272.37 g/mol}\times 100=11.75\%](https://tex.z-dn.net/?f=%5Cfrac%7Bz%5Ctimes%2016%20g%2Fmol%7D%7B272.37%20g%2Fmol%7D%5Ctimes%20100%3D11.75%5C%25)
z = 2
The molecular formula of estradiol is: ![C_xH_yO_z:C_{18}H_{24}O_2](https://tex.z-dn.net/?f=C_xH_yO_z%3AC_%7B18%7DH_%7B24%7DO_2)
Both ehhevshahahbsbdvhshs
Heat of vaporization of water will be required as water is already at it's boiling point thus heat required will be 540*10=5400 cal