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NNADVOKAT [17]
3 years ago
10

Radon-222 has a half life of 3.82 days. How much radon - 222 will remain after 11.46 days if you started with a 3.0 mg sample?

Chemistry
1 answer:
Sergio039 [100]3 years ago
8 0

Hey there!

A half-life means after a certain amount of time, half of that substance will be gone/changed after that time.

There are 3 half lives in 11.46 days because 11.46 ÷ 3.82 = 3.

So, we divide the 3mg sample in half 3 times.

3 ÷ 2 = 1.5

1.5 ÷ 2 = 0.75

0.75 ÷ 2 = 0.375

There will be 0.375mg of radon-222 remaining after 11.46 days.

Hope this helps!

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The value of the equilibrium constant K depends on: I. the initial concentrations of the reactants. II. the initial concentratio
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Answer:

None of these

Explanation:

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The equilibrium constant K is given as;

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The equilibrium constant neither depends on the concentrations of the reactants nor on that of the products.

Let us recall that at equilibrium, the concentrations of reactants and products remain largely constant. This implies that, concentration of species do not appreciably change at equilibrium because the rates of forward and reverse reactions are equal.

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The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calo
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When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

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Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

Learn more: brainly.com/question/24245395

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