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joja [24]
3 years ago
11

Draw the Lewis structure for each of the following and THEN determine the electron-pair geometry of the atom indicated. Do not i

nclude formal charges in your drawing. C in CS2 : electron-pair geometry

Chemistry
1 answer:
Hoochie [10]3 years ago
3 0

Answer: The electron pair geometry is linear

Explanation:

Carbon in carbon disulphide is surrounded by four electron pairs which is ordinarily expected to lead to a tetrahedral electron pair geometry according to Valence shell electron pair repulsion theory (VSEPR). However, the observed shape is linear due to the fact that double bonds cause more repulsion than single bonds. The double bonds force the molecule into a linear arrangement as in Carbon IV oxide molecule. See image attached for structures.

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Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine pr
denis-greek [22]

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO

Next, the corresponding moles:

C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO

Then, each element's subscripts is found to be:

C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1

Therefore, the empirical formula is:

C_4H_5N_2O

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

C_8H_10N_4O_2

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

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3 years ago
Which of the following terms best describes air? (1 point) element compound mixture none of the above
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Air is considered a homogeneous mixture. Hope that helps :)
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