What is the best chemical test for alkene?
1 answer:
You might be interested in
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: a) is the limiting reagent
b) 3.59 g
c) 11.6g
Explanation:
To calculate the moles :
According to stoichiometry :
a) 9 moles of require= 4 moles of
Thus 0.15 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
b) As 9 moles of give = 8 moles of
Thus 0.15 moles of give =
of
Mass of
c) As 9 moles of give = 3 moles of
Thus 0.15 moles of give =
of
Mass of
1.31 × 10⁴ grams.
<h3>Explanation</h3>Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:
,
where
the pressure on the gas,
;
the volume of the gas,
;
the number of moles of the gas, which needs to be found;
the absolute temperature of the gas,
.
the ideal gas constant,
if P, V, and T are in their corresponding SI units: Pa, m³, and K.
Apply the ideal gas law to find :
.
In other words, there are 410.3 moles of O₂ molecules in that container.
There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be . The mass of 410.3 moles of O₂ will be:
.
What would be the mass of oxygen in the container if the pressure is approximately the same as STP at or
instead?