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andreev551 [17]
3 years ago
15

The half-life of tritium, or hydrogen-3, is 12.32 years. After about 24.6 years, how much of a sample of tritium will remain unc

hanged?
A. 1/8
B. 1/4
C. 1/3
D. 1/2
Chemistry
2 answers:
scoundrel [369]3 years ago
8 0

Answer: The correct option is B.

Explanation: This is an example of radioactive decay and all the radioactive decay processes follow First order of kinetics.

Expression for the half life of first order kinetics is:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=12.32years

Putting in above equation, we get:

12.32=\frac{0.693}{k}\\k=0.05625year^{-1}

Expression to calculate the amount of sample which is unchanged is:

N=N_oe^{-kt}

where,

N = Amount left after time t

N_o = Initial amount

k = Rate constant

t = time period

Putting value of k = 0.05625 and t = 24.6 in above equation, we get:

N=N_oe^{-0.05625\times 24.6}

\frac{N}{N_o}=0.25

The above fraction is the amount of sample unchanged and that is equal to \frac{1}{4}

Hence, the correct option is B.

Llana [10]3 years ago
6 0
The amount of the substance left after sometime, t, is given by the equation,
                                     At = (Ai) x e^-kt
where Ai is the initial amount and k is constant. From the given half-life,
                            At / Ai = 0.5 = e^-k(12.32)      ; k = 0.5626
Then, for the next set,
                             At/Ai = e^(-0.5626)x24.6 = 
Thus, the answer is letter B.
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An object is located 51 millimeters from a diverging lens. The object has a height of 13 millimeters and the image height is 3.5
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B. 13.7 mm

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A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express
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Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

(c) Molarity = 13792 mol HNO3/m3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\  \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3

(c) HNO3 molarity (mol HNO3/m3)

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are  1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3

We can now calculate the molarity as

Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3}  =13792 \frac{molHNO3}{m3}

8 0
3 years ago
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