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Hunter-Best [27]
3 years ago
8

Iodine-131, t1/2 = 8.0 days, is used in the diagnosis and treatment of thyroid gland diseases. If a laboratory sample of iodine-

131 initially emits 9.95 × 1018 β particles per day, how long will it take for the activity to drop to 6.22 × 1017 β particles per day?
Chemistry
1 answer:
SashulF [63]3 years ago
7 0

Answer:

32,0 days.

Explanation:

The radioactive decay follows:

N_{t} = N_{0}e^{\frac{-0.693t}{t_{1/2}}

Where Nt is the concentration in a time t (6,22x10¹⁷), N₀ is the initial concentration (9,95x10¹⁸) Half life time is 8,0 days and t is the time it take to drop this concentration. Replacing:

6.22x10^{17} = 9,95x10^{18}e^{\frac{-0.693t}{8 days}

0,0625 = e^{\frac{-0.693t}{8days}

ln 0,0625 = {\frac{-0.693t}{8days}

-2,77*8days = -0.693t

-22,2days = -0.693t

32,0days = t

It take <em>32,0 days</em>

<em></em>

I hope it helps!

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A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>

1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

3 0
3 years ago
Groundwater in an area flows at a speed of 4km/h. How long would it take the water to flow 10 km t it’s springs
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Explanation:

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Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
yaroslaw [1]

Answer:

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

Explanation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

initially

3.0 atm      0              0

At equilibrium

(3.0-2p)     p               3p

Equilibrium partial pressure of nitrogen gas = p = 0.90 atm

The expression of a pressure equilibrium constant will be given by :

K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}

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12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

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3 years ago
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