The following are electronegativity. What value should be where the M is?0.981.57M2.553.043.443.98No data

Chemistry

1 answer:

ipn [44]1 year ago

6 0

Answer 2.04

Explanation

Electronegativity decreases down the group and increases across the period in the periodic table.

In the period two of the periodic table,we have the following values for electronegativities with respect to its elements.

Li...........0.98

Be.........1.57

B..........2.04

C.........2.55

N..........3.04

O...........3.44

F...........3.98

Ne........n.a

The value that should be where M is is 2.04

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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:

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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.

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

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

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