Answer:

Explanation:
Given that:
The Half-life of
=
is less than that of 
Although we are not given any value about the present weight of
.
So, consider the present weight in the percentage of
to be y%
Then, the time elapsed to get the present weight of
= 
Therefore;

here;
= Number of radioactive atoms relating to the weight of y of 
Thus:

--- (1)
However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of
to be = 
Then:
---- (2)
here;
= Number of radioactive atoms of
relating to 3.0 a/o weight
Now, equating equation (1) and (2) together, we have:

replacing the half-life of
=
( since
)
∴

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o
Thus, The time elapsed is 
Answer:
Explanation:
When comparing the drops of oil and water, one thing I noticed was the shape. The water drop was more defined, whereas the drop of oil began to spread and was much flatter. This may be due to the waxy material, and how both oil and water react to the wax.
Explanation:
Let us assume that the given data is as follows.
V = 3.10 L, T =
= (19 + 273)K = 292 K
P = 40 torr (1 atm = 760 torr)
So, P = 
= 0.053 atm
n = ?
According to the ideal gas equation, PV = nRT.
Putting the given values into the above equation to calculate the value of n as follows.
PV = nRT

0.1643 = 
n = 
It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.
No. of moles =
mass =
g
= 0.315 g
Thus, we can conclude that the mass of liquid ethanol is 0.315 g.
Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>