Juliette uses 20 N to move her desk 5 meters. Which of the following is true?

A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

6 0

2 years ago

When 100. mL of 6.00 M HCl is diluted to 300. mL, the final concentration is ________.

Marat540 [252]

Answer:

2.00 M

Explanation:

The concentration of a solution is given by

$M=\frac{m}{V}$

where

m is the mass of solute

V is the volume of the solution

At the beginning, the solution has:

M = 6.00 M is the concentration

V = 100 mL = 0.1 L is the volume

So the mass of solute (HCl) is

$m=MV=(6.00)(0.1)=0.6g$

Then, the HCL is diluted into a solution with volume of

V = 300 mL = 0.3 L

Therefore, the final concentration is:

$M=\frac{m}{V}=\frac{0.6}{0.3}=2.00 M$

3 0

3 years ago

Read 2 more answers

The mass percent of carbon in a typical human is 18%, and the mass percent of 14C in natural carbon is 1.6 × 10-10%. Assuming a

Tema [17]

Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

$\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g$

Carbon-14 is $1.6\times 10^{-10}\ \%$ of the carbon in the body. So,

$\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g$

Also,

14 g of Carbon-14 contains $6.023\times 10^{23}$ atoms of carbon-14

So,

$2.48\times 10^{-8}\ g$ of Carbon-14 contains $\frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8}$ atoms of carbon-14

Atoms of carbon-14 = $1.07\times 10^{15}$

Given that:

Half life = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.00012 years⁻¹

Also, 1 year = $3.154\times 10^7$ s

So, The rate constant, k = $\frac{0.00012}{3.154\times 10^7}$ s⁻¹ = $3.8\times 10^{-12}\ s^{-1}$

Thus, decay events per second = $K\times atoms decayed$ = $3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s$ = 4066 decay/s

4 0

3 years ago

Which of the following is a characteristic of both the Earth and the Moon?

4vir4ik [10]

Answer:

rocky surface covered with landforms

Explanation:

7 0

3 years ago

Three Stoichiometry Questions

andrezito [222]

Answer:

Explanation:

7)

Given data:

Mass of aluminium = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 27 g/mol

Number of moles = 0.09 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 32 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of aluminium oxide with aluminium and oxygen.

Al ; Al₂O₃

4 : 2

0.09 : 2/4×0.09 = 0.045

O₂ : Al₂O₃

3 : 2

0.08 : 2/3 ×0.08 = 0.053

The number of moles of aluminium oxide produced by Al are less so it will limiting reactant.

Mass of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.045 × 101.96 g/mol

Mass = 4.6 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.5 g / 4.6 ×100

Percent yield = 76.1%

8)

Given data:

Mass of copper produced = 3.47 g

Mass of aluminium = 1.87 g

Percent yield = ?

Solution:

Chemical equation:

2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 1.87 g/ 27 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of copper with aluminium.

Al ; Cu

2 : 3

0.07 : 3/2×0.09 = 0.105

Mass of copper:

Mass = number of moles × molar mass

Mass = 0.105 × 63.55 g/mol

Mass = 6.67 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.47 g / 6.67 × 100

Percent yield = 52%

4 0

3 years ago