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3 years ago

In an electron-dot structure of an element, the dots are used to represent ________.

Chemistry

1 answer:

Vlad [161]3 years ago

3 0

Dots are used to represent electrons.

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A student uses a sample of KOH stock solution and dilutes it to a total of 120 mL. If the diluted solution is 0.60 M KOH and its

Vesnalui [34]

Answer:

5.5

Explanation:

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3 years ago

After taking a hot shower you may notice that the room gets steamy and the mirror is foggy you might also notice that there is m

ruslelena [56]

Answer:the temperature might increase because of the humidity in the room so everything gets goofy and moist

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3 years ago

What is the significant digit in 23.45

klio [65]

If you mean the number of Significant Figures/Digits in 23.45, it would be 4. This is because every single non-zero digit is counted as a significant figure

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3 years ago

An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the

san4es73 [151]

Answer:

the ionic radius of the anion $r^- = 312.52 \ pm$

Explanation:

From the diagram shown below :

The anion $Cl^-$ is located at the corners

The cation $Cs^+$ is located at the body center

The Body diagonal length = $\sqrt{3 \ a }$

∴ $2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a$

Given that :

$\frac{r^+}{r^-} =0.84$ (i.e the ratio of the ionic radius of the cation to the ionic radius of

the anion )

$0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a$

Also ; a = 664 pm

Then :

$r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm$

Therefore, the ionic radius of the anion $r^- = 312.52 \ pm$

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3 years ago

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

$\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\$

$\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\$

$\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\$

$\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}$

$E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\$

$E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}$

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3 years ago