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torisob [31]
3 years ago
10

Americium-241 is an alpha emitter used in household smoke detectors and K-40 is a positron emitter used to date rock samples. Th

e daughter nuclides in these two reactions are:
Chemistry
2 answers:
natita [175]3 years ago
5 0

Answer:

Neptunium - 237 , Argon - 40 .

Explanation:

Americium - 241   ⇒    Neptunium - 237 + alpha particles

²⁴¹₉₅Am    ⇒     ²³⁷₉₃Np  + ⁴₂α

K-40   ⇒  Argon 40 + positron .

⁴⁰₁₉K    ⇒   ⁴⁰₂₀Ar   +   ₁e

VMariaS [17]3 years ago
4 0

Answer:

Np-237, Ar-40.

Explanation:

Alpha particle is positively charged particle having a mass of 4 unit. Hence,when alpha is  removed from any isotope it produces a element having  mass 4 units less than original isotope that was before emission. Americium 241 isotopes used in the smoke detector which emit an alpha particle to produce neptunium-237.

Potassium-40 emits positron which is having negative charge to lower one proton and increase one neutron number in the particle and in such way it will produce ^{40}Ar_{18} not  ^{40}Ca_{20}

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Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
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<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

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We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

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<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

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<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

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Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
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Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

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