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snow_lady [41]
3 years ago
8

assume you have a 500 g sample of a radioactive isotope with a half-life of 100 years. fill in the table showing how many grams

of material will remain after the indicated time has passed.
Chemistry
2 answers:
djverab [1.8K]3 years ago
6 0
8.10 hours is the answer
sweet-ann [11.9K]3 years ago
3 0

Answer:

8.10 hours.

Explanation:

You start with 500.0g.

After the first half-life, you have 250.0g.

After the second, you have 125.0g.

After the third, you have 62.50g.

Therefore, it takes three half-lives to decay to 62.50g.

Therefore, the elapsed time must be triple the length of one half-life.

24.3

3

=

8.10

, so it is 8.10 hours.

Explanation:

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Reptile [31]

 The half  life  of uranium- 238  is

 4.46 x10^9  years


 Explanation

Half life  is the time  taken for  the  radioactivity  of   a isotope  to fall  to half  its  original  value.

The  original   mass  of uranium-238  is  4.0 mg

  Half  of original  mass of uranium  = 4.0  mg /2 = 2.0  mg

since  it  take 4.46  x 10^ 9 years  for  the sample   to half  the half life of uranium -238 = 4.46 x10^9 years

7 0
3 years ago
Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i
anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

K=\frac{Products}{reagents}

In any case  

aA +Bb  equilibrium Cd +dD

where K is:

K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})

Where: H is reaction enthalpy, R is the gas constant and T temperature.  

Clearing the equation for K_{2} we get:

K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

Here we can study two cases: when delta H^{o} is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so K_{2} is greater that K_{1} and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so K_{2} will be less than K_{1} the forward reactions is not favored.  

{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

5 0
3 years ago
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