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Phantasy [73]
3 years ago
13

Consider a certain type of nucleus that has a half-life of 32 min. calculate the percent of original sample of nuclides remainin

g after 1.9 hours have passed
Chemistry
1 answer:
Irina18 [472]3 years ago
8 0
t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.

32 = 0.693 / λ 
λ   = 0.693 / 32          (1) 

Nt = Nο eΛ(-λt)          (2)

Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min

From (1) and (2),


Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt =  0.085Nο 

Percentage = (Nt/Nο) x 100%
                   = (0.085Nο/Nο) x 100%
                   = 8.5%

Hence, Percentage of remaining atoms with the original sample is 8.5%

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3 years ago
When the moon moves directly between the sun and Earth and casts its shadow over part of Earth, you are
chubhunter [2.5K]

Answer:C

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7 0
2 years ago
Read 2 more answers
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
Which types of orbitals are found in the principal energy level n = 2?
AfilCa [17]
Remember this:

1) n is principal quantum number and represents the energy level.

2) l is the second quantum number and represent the type of orbital.

3) l can take values from 0 to n - 1

4) each number of l is associated with a type of orbital.  This table shows the equivalence:
 
l number       type of orbital

0                    s

1                    p

2                    d

3                     f

With that, you can tell that n = 2 permits l = 0 and 1, which is orbitals s and p.

Therefore, the answer is the option D) s, p.
3 0
3 years ago
Read 2 more answers
at 55 years what mass remains of a 200.0 g sample of a radioactive isotope with a half life of 10.0 years
Harman [31]

Answer:

4.419 g

Explanation:

     55 years is 5.5 half lives

200 g   *   (1/2)^5.5 = 4.419 g

3 0
1 year ago
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