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m_a_m_a [10]
3 years ago
8

After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?

Chemistry
1 answer:
Bond [772]3 years ago
5 0
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value.  The equation to describe the decay is
Nt=N0(1/2) ^{t/t(1/2)}  where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time.  So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
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Varvara68 [4.7K]

Answer:

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hope it helps

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3 0
2 years ago
Read 2 more answers
7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are
Juli2301 [7.4K]

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] =  0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

6 0
3 years ago
Read 2 more answers
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Advocard [28]
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