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3 years ago

2. Mercury-197 is used for kidney scans and has a half-life of 3 days. If 32 grams of

Chemistry

1 answer:

VLD [36.1K]3 years ago

8 0

<h3>Answer:</h3>

1 grams Mercury-197

<h3>Explanation:</h3>

From the question,

The radioisotope is Mercury-197

The half life of Mercury-197 is 3 days

Starting mass is 32 grams

Time taken is 15 days

We are required to calculate the final mass of the radioisotope after 15 days.

Note that, half life is the time taken by radioisotope to decay by half its original amount.
In this case, the half life of mercury-197 is 3 days, therefore, it takes three days for a mass of mercury-197 to decay by half its original mass.

Therefore, using the formula;

Final amount = Starting mass × (1/2)^n

where n is the number of half-lives (ratio of time to half life)

In our case, n = 15 days ÷ 3 days

= 5

Therefore;

Final mass = 32 g × (1/2)^5

= 32 g × 1/32

= 1 g

Therefore, the amount of mercury-197 in the shipment that will arrive is 1 g.

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Ivahew [28]

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We'll begin by calculating the number of mole of in 0.16 g of Ca(OH)₂. This can be obtained as follow:

Mass of Ca(OH)₂ = 0.16 g

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Mole = mass / molar mass

Mole of Ca(OH)₂ = 0.16 / 74

Next, we shall determine the molarity of the stock solution of Ca(OH)₂.

Mole of Ca(OH)₂ = 0.00216 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

<h3>Molarity of Ca(OH)₂ =? </h3>

Molarity = mole / Volume

Molarity of Ca(OH)₂ = 0.00216 / 0.1

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Next, we shall determine the molarity of the diluted solution. This can be obtained as follow:

Volume of stock solution (V₁) = 100 mL

Molarity of stock solution (M₁) = 0.0216 M

Volume of diluted solution (V₂) = 250 mL

<h3>Molarity of diluted solution (M₂) =?</h3>

0.0216 × 100 = M₂ × 250

2.16 = M₂ × 250

Divide both side by 250

M₂ = 2.16 / 250

Thus, the molarity of the diluted solution is 0.00864 M

Finally, we shall determine the concentration of the hydroxide ions, OH¯ in the diluted solution. This can be obtained as follow:

Ca(OH)₂(aq) —> Ca²⁺(aq) + 2OH¯(aq)

From the balanced equation above,

1 mole of Ca(OH)₂ contains 2 moles of OH¯

Therefore,

0.00864 M Ca(OH)₂ will contain = 2 × 0.00864 = 0.01728 M OH¯

Thus, the concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

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