CHEMISTRY HIGH SCHOOL

Samarium-146 has a half-life of 103.5 million years. After 1.035 billion years, how much samarium-146 will remain from a 205-g sample?0.200 g
0.400 g
20.5 g
103 g

Answers

Answer 1
Answer:

Answer : The correct option is, 0.200 g

Solution :

As we know that the radioactive decays follow first order kinetics.

First we have to calculate the rate constant of a samarium-146.

Formula used :

t_(1/2)=(0.693)/(k)

Putting value of t_(1/2) in this formula, we get the rate constant.

103.5* 10^6=(0.693)/(k)

k=6.6* 10^(-9)year^(-1)

Now we have to calculate the original amount of samarium-146.

The expression for rate law for first order kinetics is given by :

k=(2.303)/(t)\log(a)/(a-x)

where,

k = rate constant  = 6.6* 10^(-9)year^(-1)

t = time taken for decay process  = 1.035* 10^(9)years

a = initial amount of the samarium-146 = 205 g

a - x = amount left after decay process  = ?

Putting values in above equation, we get the value of initial amount of samarium-146.

6.6* 10^(-9)=(2.303)/(1.035* 10^(9))\log(205)/(a-x)

a-x=0.200g

Therefore, the amount left of the samarium-146 is, 0.200 g
Answer 2
Answer:

Ans: 0.200 g

Given:

Half life of Sm-146 = t1/2 = 103.5 million years

Time period, t = 1.035 billion years = 1035 million years

Original mass of sample, [A]₀ = 205 g

To determine:

Amount of sample after t = 1035 million years

Explanation:

The rate of radio active decay is given as:

A(t) = A(0)e^(-0.693t/t1/2) \n\n= 205 g * e^{(0.693*1035)/(103.5) } \n\n= 0.200 g

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Answers

For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

                                ΔTf = (Kf)(m)(i)
where: 
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1246

Answers

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Answer:

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Explanation:

I got it right on edge

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Answers

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Answers

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Answers

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