Radioactive decay => C = Co { e ^ (- kt) |
Data:
Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min
Time conversion: 4 hr 39 min = 4.65 hr
1) Replace the data in the equation to find k
C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}
=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719
2) Use C / Co = 1/2 to find the hallf-life
C / Co = e ^ (-kt) => -kt = ln (C / Co)
=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k
t = ln(2) / 0.44719 = 1.55 hr.
Answer: 1.55 hr
According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m
Answer:
In the unfolded polypeptide, there are ordered solvation shells of water around the protein
groups. The number of water molecules involved in such ordered shells is reduced when the protein
folds, resulting in higher entropy. Hence, the lower free energy of the native conformation.
Explanation:
