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2 years ago

14

Initially a beaker contains 225.0 mL of a 0.350 M MgSO4 solution. Then 175.0 mL of water are added to the beaker. Find the conce

ntration of the final solution

2 answers:

Rasek [7]2 years ago

8 0

Answer:

400.0

Explanation:

I just got it right

givi [52]2 years ago

7 0

Answer:

came

Explanation:

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worty [1.4K]

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2 years ago

The half-life for the reaction below was determined to be 2.14 × 10^4 s at 800 K. The value of the half-life is independent of t

lord [1]

Answer:

0.0907 s

Explanation:

This an Arrhenius equation problem, so you relate the half-life with the kinetic constant of the reaction in order to calcule the same thermodynamic parameters at another temperature.

To calcule the kinetic constant of the reaction you need to know the order of it, look closely to the sentence "The value of the half-life is independent of the inital concentration of N2O present." the only order independent from the initial concentration of reagents is first order, so you can calculate K at 800 K, using:

$k(800 K)=\frac{ln(2)}{t_{1/2}}= \frac{ln(2)}{2.14 * 10^{4} s}}=3.239*10^{-5}s^{-1}}$

Now you can use Arrhenius equation to calcule K at 1150.66 K

$ln(\frac{k1}{k2} )=-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} )$

$k2= k1*exp(-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} ))=3.239*10^{-5}s^{-1}*exp(-\frac{270 000 J/mol }{8.314 J/mol *k }(\frac{1}{1150.66K} - \frac{1}{800K} ))=7.639 s^{-1}$

Then calculate the new half-life:

$t_{1/2} =\frac{ln(2)}{7.639s^{-1}}=0.0907 s$

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3 years ago