The following Lewis structures represent valid resonance forms. true or false
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:
To find the number of moles of each element present in 1 mole of Be(OH)₂, we use the ratio of the molar masses of the elements to the compound and multiply by the given mole.
Molar mass of Be(OH)₂ = 9 + 2(16+1) = 43g/mol
number of moles of Be = x 1 = 0.209mole
number of moles of O = x 1 = 0.744mole
number of moles of H = x 1 = 0.047mole