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tensa zangetsu [6.8K]
3 years ago
6

The half-life of a radioactive kind of strontium is 3 hours. If you start with 64 grams of it, how

Chemistry
1 answer:
aliina [53]3 years ago
7 0

Answer:

8

Explanation:

If strontium has a half-life of 3 hours and you have 64 grams of it that means every 3 hours strontium's grams will be divided by 2.

So you have 64/2 = 32 for the first half life   {Time is currently at 3 hours}

32/2 = 16 for the second half life {Time is currently 6 hours}

Then you have 16/2 = 8 for the third half-life {Time is currently 9 hours}

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Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surrou
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We know,

\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol

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\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol

For , 2.41 moles of I_2 :

\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ

We know :

\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K

Hence, this is the required solution.

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