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Elden [556K]
3 years ago
8

A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24

has a half-life of 15 h what is the activity after each of the following intervals?
a. one half-life
b. 30h
c. three half-lives
d. 2.5 days
Chemistry
1 answer:
SOVA2 [1]3 years ago
5 0

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

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<h3>What is a data set?</h3>

Generally, Simply put, data sets are groups of information. When dealing with tabular information, a data set is analogous to one or more database tables, where each column represents a different variable and each row represents a separate record.

In conclusion, the total amount of numbers that are either more than or equal to 8

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4 0
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Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
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Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

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I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
2 years ago
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