In is a lot easier because it uses 10s.
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Answer:
the value of equilibrium constant for the reaction is 8.5 * 10⁷
Explanation:
Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)
equilibrium constant Kc = ![\frac{1}{[Cl_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BCl_2%5D%5E2%7D)
Given that,
We are given:
Equilibrium amount of titanium = 2.93 g
Equilibrium amount of titanium tetrachloride = 2.02 g
Equilibrium amount of chlorine gas = 1.67 g
We calculate the No of mole = mass / molar mass
mass of chlorine gas = 1.67 g
Molar mass of chlorine gas = 71 g/mol
mole of chlorine = 1.67 / 71
= 7.0L
Concentration of chlorine is = no of mole / volume
= 0.024 / 7
= 3.43 * 10⁻³M
equilibrium constant Kc = ![\frac{1}{[Cl_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BCl_2%5D%5E2%7D)
= ![\frac{1}{[3.43 * 10^-^3]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5B3.43%20%2A%2010%5E-%5E3%5D%5E2%7D)
= 8.5 * 10⁷
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g