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2 years ago

If the half life of a radioactive isotope is 1 day, then how much of the original isotope remains at the end of two days?

Chemistry

2 answers:

ikadub [295]2 years ago

8 0

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25%

Explanation:

In a half life of a radioactive isotope is 1 day,it means it loses its half mass each day

We a formula for N half life

$\sf { \frac{1}{2} }^{n}$

where n is the number of days

Here the isotope is kept for 2 days

so it's left over mass will be

$\sf { (\frac{1}{2}) }^{2} \\ \\ \\ \frac{1}{2} \times \frac{1}{2} \\ \\ \sf \frac{1}{4}$

It's left over mass 1/4th of the original mass

Now, we need to find it's percentage by multiplying with 100

$\sf \frac{1}{4} \times 100 \%\\ \\ \sf \cancel\frac{100}{4} \% \\ \\ \sf 25\%$

So 25% mass will be left after 2 day of half life radioactive isotope.

JulijaS [17]2 years ago

8 0

25% because the half life of 100 is 50 and the half life of 50 is 25

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Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine

Marrrta [24]

Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

Firstly, we should write the balanced equation of the reaction:

N₂ + 3H₂ → 2NH₃.

1) To determine the limiting reactant of the reaction:

From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.

This means that N₂ reacts with H₂ with a ratio of (1:3).

We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation: n = mass / molar mass.

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

2) To determine the amount (in grams) of excess reactant of the reaction:

As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.

∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

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zavuch27 [327]

Answer:

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Explanation:

To solve this problem you can use Clausius-Clapeyron equation that serves to estimate vapor pressures or temperatures:

$Ln(\frac{P_{2}}{P_{1}} ) =\frac{ deltaH_{vap}}{R} (\frac{1}{T_{1}}-\frac{1}{T_{2}} )$

Where:

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R is gas constant 8,314x10⁻³ kJmol⁻¹K⁻¹

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T2 is 54,81°C + 273,15 = 327,96 K

Thus:

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Thus, P2 is 2,54x10² mmHg

I hope it helps!

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