Answer:
19 years
Explanation:
Tritium follows a first-order decay which can be represented by the following expression.
![ln(\frac{[H]_{t}}{[H]_{0}} )=-k.t](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BH%5D_%7Bt%7D%7D%7B%5BH%5D_%7B0%7D%7D%20%29%3D-k.t)
where,
[H]t is the concentration of tritium at a certain time t
[H]₀ is the initial concentration of tritium
k is the rate constant
If we know the half-life (t1/2), we can calculate the rate constant.
![k=\frac{ln2}{t_{1/2}} =\frac{ln2}{12.3y} =0.0564y^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln2%7D%7Bt_%7B1%2F2%7D%7D%20%3D%5Cfrac%7Bln2%7D%7B12.3y%7D%20%3D0.0564y%5E%7B-1%7D)
![ln(\frac{[H]_{t}}{[H]_{0}} )=-k.t\\ln(\frac{0.34[H]_{0}}{[H]_{0}} )=-(0.0564y^{-1}).t\\t=19y](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BH%5D_%7Bt%7D%7D%7B%5BH%5D_%7B0%7D%7D%20%29%3D-k.t%5C%5Cln%28%5Cfrac%7B0.34%5BH%5D_%7B0%7D%7D%7B%5BH%5D_%7B0%7D%7D%20%29%3D-%280.0564y%5E%7B-1%7D%29.t%5C%5Ct%3D19y)
Mass of Calcium phosphate : = 21127.7 g
<h3>Further explanation</h3>
Given
4.1 x 10²⁵ particles of Calcium phosphate
Required
mass
Solution
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
mol Calcium phosphate :
n = N : No
n = 4.1 x 10²⁵ : 6.02 x 10²³
n = 68.11
mass Calcium phosphate Ca3(PO4)2 :
= mol x MW
= 68.11 x 310.2 g/mol
= 21127.7 g
= 21.13 kg
Answer:Calculations must be done first because energy depends on both temperature and mass.
Explanation:
first we find the mass of the final solution
we know that
density=mass/volume
we know the density and volume
1=m/100
m=100g
now we calculate the mass ofNa2CO3
2% of 100 is equal with 2g of Na2CO3
Answer: pH = 6.77
Explanation:
1) <u>Chemical equilibrium</u>
- 2 H₂O (l) ⇄ H₃O⁺ (aq) + OH⁻ (aq)
2) <u>Equilibrium constant, Kw</u>
- By stoichiometry [H₃O⁺] = [OH⁻]. Call it x
- x = √ (2.92 × 10⁻¹⁴) = 1.709 × 10⁻⁷ M = [H₃O⁺]
3)<u> pH</u>
- pH = - log [H₃O⁺] = - log (1.709 × 10⁻⁷) = 6.77