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LenaWriter [7]
1 year ago
12

Moles of C2H6 in 3.75x10 ^23 molecules of C2H6

Chemistry
2 answers:
Blizzard [7]1 year ago
5 0

Answer : The number of moles of C_2H_6 is, 0.622 moles

Solution :

As we know that,

1 mole contains 6.022\times 10^{23} number of molecules.

As, 6.022\times 10^{23} number of molecules of C_2H_6 present in 1 mole of C_2H_6

So, 3.75\times 10^{23} number of molecules of C_2H_6 present in \frac{3.75\times 10^{23}}{6.022\times 10^{23}}=0.622 mole of C_2H_6

Therefore, the number of moles of C_2H_6 is, 0.622 moles

Elan Coil [88]1 year ago
3 0
To determine the moles of C2H6 from the number of molecules, we use the avogadro's number to relate moles into molecules. From avogadro's number, 1 mol is equal to 6.022x10^23 molecules. 

<span> 3.75x10 ^23 molecules C2H6 (1 mol / 6.022x10^23 molecules ) = 0.6227 mol C2H6

Hope this answers the question.</span>
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Answer :

(1) The number of valence electrons present in the compound is, 20

(2) The number of bonded electrons present in the compound is, 16

(3) The number of lone pair electrons present in the compound is, 4

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Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

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Therefore, the total number of valence electrons in CH_3CH_2OH = 2(4) + 6(1) + 6 = 20

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The Lewis-dot structure of CH_3CH_2OH is shown below.

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1 year ago
What is the axmen classification for benzene (C6H6)?
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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

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Best regards.

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Answer:

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