Hydrofluoric acid is what type of acid?
1 answer:
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Given: Half-life of <span>133Ba = t1/2 = 10.5 years.
The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.
Also, </span>
where, Co = initial concentration = <span>1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................
Answer 1: For t = </span><span>5 years
</span>
Therefore,
= 0.1432
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.
Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................
Answer 2: For t = 30 years
Therefore, = 0.8597
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.
Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................
Answer 3: t = 180 years
Therefore, = 5.1585
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.
Number of 133Ba atoms left after 180 years = 7.6367 X10^4.
The radio-active materials obeys 1st order dissociation kinetics. Therefore we have:
k = 0.693 / t1/2 = 0.693 / 10.5 = 0.066 years-1.
Also, </span>
where, Co = initial concentration = <span>1.1×10^10 atoms
Ct = conc. of Ba at time t.
......................................................................................................................
Answer 1: For t = </span><span>5 years
</span>
Therefore,
Therefore, Co/Ct = 1.3908
Therefore, Ct = 7.9086 X 10^9 atoms.
Number of 133Ba atoms left after 5 years = 7.9086 X 10^9.
....................................................................................................................
Answer 2: For t = 30 years
Therefore,
Therefore, Co/Ct = 7.2402
Therefore, Ct = 1.5193 X 10^9 atoms.
Number of 133Ba atoms left after 30 years = 1.5193 X 10^9.
........................................................................................................................
Answer 3: t = 180 years
Therefore,
Therefore, Co/Ct = 1.44 X 10^5
Therefore, Ct = 7.6367 X10^4 atoms.
Number of 133Ba atoms left after 180 years = 7.6367 X10^4.
Answer:
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte like urea)
= freezing point constant =
m= molality
Weight of solvent (X)= 950 g = 0.95 kg
Molar mass of non electrolyte (urea) = 60.06 g/mol
Mass of non electrolyte (urea) added = ?
Thus urea was dissolved.