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Neko [114]
3 years ago
8

7. Phosphorus 32 ( 32), a radioactive isotope of phosphorus, has a half-life of 14.2 days. a. If the amount of radioactivity of

phosphorus 32 decays continuously at a constant rate, what is the decay constant? Round your answer to 4 decimal places. (10 marks) If 100 grams of this substance are present initially, determine the exponential decay function by using the decay constant found in part (a). (4 marks) c. What amount of radioactivity will be left after 6 days? Round your answer to 2 decimal places. (4 marks)
Chemistry
1 answer:
Wittaler [7]3 years ago
4 0

Explanation:

(a) Given that:

Half life = 14.2 days

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

Where, k is rate constant

So,  

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

k=\frac {ln\ 2}{14.2}\ days^{-1}

<u>The rate constant, k = 0.0488 days⁻¹ </u>

(b)Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration  = 100 g

So,  

[A_t]=100\times e^{-0.0488\times t}\ g

(c) Given t = 6 days

So,

[A_t]=100\times e^{-0.0488\times 6}\ g

<u>Amount left = 74.6171 g</u>

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Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

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= 2 × -4.6 × 10³kJ / -1.65×10³kJ

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Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

#SPJ4

DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

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