Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
Use the equation m1 * v1 = m2 * v2
Example: (2.75 L)(1.5 M) = (12.0 M)(v2) then solve for v2
You may have to solve for m2 as well just plug and chug
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