Answer:
2.1x10⁹ years
Explanation:
U-238 is a radioactive substance, which decays in radioactive particles. It means that this substance will lose mass, and will form another compound, the Pb-206.
The time need for a compound loses half of its mass is called half-life, and knowing the initial mass (mi) and the final mass (m) the number of half-lives passed (n) can be found by:
m = mi/2ⁿ
The mass of Pb-206 will be the mass that was lost by U-238, so it will be mi - m. Thus, the mass ration can be expressed as:
(mi-m)/m = 0.337/1
mi - m = 0.337m
mi = 1.337m
Substituing mi in the expression of half-life:
m = 1.337m/2ⁿ
2ⁿ = 1.337m/m
2ⁿ = 1.337
ln(2ⁿ) = ln(1.337)
n*ln(2) = ln(1.337)
n = ln(1.337)/ln2
n = 0.4190
The time passed (t), or the age of the sample, is the half-life time multiplied by n:
t = 4.5x10⁹ * 0.4190
t = 1.88x10⁹ ≅ 2.1x10⁹ years
1 gram = 1000 mg
=>
1 mg = 1/1000 g
25 mg = 25* (1/1000) g = 25/1000 g = 0.025g
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
