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WITCHER [35]
2 years ago
12

If the amount of radioactive sodium-24, used for studies of electrolytes within the body, in a sample decreases from 0.8 to 0.2

mg in 30.0 h , what is the half-life of sodium-24?
Chemistry
1 answer:
kirill [66]2 years ago
6 0

Answer:

15 h

Explanation:

Okay, the first thing that we all have to know before we can answer this question is that this Topic that is, Chemistry of Radioactivity is related to kinetics in a way that Radioactive disintegration follows the first order of Reaction which is under kinetics. So, we will be using the first order kinetics rate law to answer this question. Using the equation (1) below;

k =[ 2.303/ t ]×log ([N°}/ [Nr]) --------(1).

We are given from the question that N° = initial sample = 0.8 mg and Nr= sample remaining = 0.2 and the time taken = t= 30.0 h.

k= (2.303/ 30.0 h ) × log (0.8/0.2).

k=0.076768 h^-1 × log (4).

k= 0.076768 h^-1 × 0.6021.

k= 0.0462 h^-1.

Therefore, using the formula for Calculating half life below for first order kinetics we will be able to find out answer.

k = ln 2/ t(1/2). Where t(1/2) is the half life.

t(1/2) = ln 2/ k.

t(1/2) = ln 2 / 0.0462 h^-1.

t(1/2)= 0.6931/0.0462 h^-1.

t(1/2)=15 h

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Answer:

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The potentiometric pH meter comprises a pair of electrodes and a basic electronic amplifier, some may even comprise a combination electrode and some sort of display that demonstrates pH units. The potentiometric pH meter generally exhibits a reference electrode or a combination electrode, and a glass electrode. The probes or electrodes are administered within a solution whose pH values are needed to be determined.  

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The correct option is: <u>B. 366 torr</u>

Explanation:

Given: <u>On the ground</u>- Initial Volume: V₁ = 8.00 m³, Initial Atmospheric Pressure: P₁= 768 torr;

<u>At 4200 m height</u>- Final Volume: V₂ = 16.80 m³, Final Atmospheric Pressure: P₂ = ?

Amount of gas: n, and Temperature: T = constant

<u>According to the Boyle's Law</u>, for a given amount of gas at constant temperature:         P₁ V₁ = P₂ V₂

⇒  P₂ = P₁ V₁ ÷ V₂

⇒  P₂ = [(768 torr) × (8.00 m³)] ÷ (16.80 m³)

⇒  P₂ = 365.71 torr ≈ 366 torr

<u>Therefore, the final air pressure at 4200 m height: P₂ = 366 torr.</u>

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