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DENIUS [597]
3 years ago
6

Which mathematical statement correctly relates qsys and qsurr? A. qsys = qsurr B. qsys < qsurr C. qsys > –qsurr D. qsys =

–qsurr
Chemistry
1 answer:
tatiyna3 years ago
4 0

Answer:

  • <u><em>Option D. qsys = - qsurr</em></u>

Explanation:

The symbol q is used to denote heat energy.

Considering positive the heat absorbed and negative the heat released:

  • <em>qsys</em> is the heat absorbed by the systme

  • <em>qsurr</em> is the heat absorbed by the surroundings

When the system does not do work on or receive work from the surroundings, the first law of thermodynamics states that:

  • <em>qsys </em>+ <em>qsurr</em> = 0

From which:

  • <em>qsys = - qsurr ← </em>answer

That is the option D.

That means that, the heat abosorbed by the system (if <em>qsys</em> is positive) equals the heat released by the surroundings, or the heat released by the system (if <em>qsys</em> is negative) equals the heat absorbed by the surroundings.

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If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?
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<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

k = rate constant = ?

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[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

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Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

Hence, the half life of the sample of silver-112 is 3.303 hours.

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The other factor will not be improved that is its reliability and precision remains the same.

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