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3 years ago

Hurry PLEASE HELP!

Chemistry

1 answer:

avanturin [10]3 years ago

7 0

B. 11,540

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Nt=25 g

No=100 g

t1/2=5770 years

$\tt 25=100\dfrac{1}{2}^{T/5770}\\\\\dfrac{1}{4}=\dfrac{1}{2}^{T/5770}\\\\2=T/5770\rightarrowT=11540~years$

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aliina [53]

Answer:

<h2>The answer you are looking for is (B)</h2>

Explanation:

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8 0

3 years ago

Read 2 more answers

2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)

defon

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity = $\frac{number of moles}{volume}$

Solve for number of moles

Number of moles = Volume x Molarity

Substitution

Number of moles = 100 x 0.000013

Number of moles = 0.0013

2.- Calculate the mass of Pb formed.

207.2 g of Pb ----------------- 1 mol

x g ----------------- 0.0013 moles

x = (0.0013 x 207.2) / 1

x = 0.269 g of Pb

8 0

3 years ago

Read 2 more answers

A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay fo

Ksenya-84 [330]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$80 = x*20$

$80 = 20\:x$

$20\:x = 80$

$x = \dfrac{80}{20}$

$\boxed{x = 4}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{43}{2^{4}}$

$m = \dfrac{43}{16}$

$\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

8 0

3 years ago

Iodine-131 is a beta emitter used as a tracer in radio immunoassays in biological systems. It follows first order kinetics. The

allsm [11]

<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

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

where,

$t_{1/2}$ = half life of the reaction = 8.04 days

Putting values in above equation, we get:

$k=\frac{0.693}{8.04days}=0.0862days^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.0862days^{-1}$

t = time taken for decay process = 39 days

$[A_o]$ = initial amount of the sample = 8.0 grams

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}$

$[A]=0.278g$

Hence, the amount of Iodine-131 remain after 39 days is 0.278 grams

3 0

3 years ago

Its silver tarnishing physical or chemical change.

Mrac [35]

Yes it is a chemical change because it is changing color which shows a chemical change, plus it changes into another element

3 0

3 years ago