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1 year ago

A compound with the formula C6H12 may or may not be a saturated hydrocarbon.

Chemistry

1 answer:

astraxan [27]1 year ago

6 0

A compound with the formula C6H12 is not considered a Saturated hydrocarbon.

Why is C6H12 isn't considered a Saturated hydrocarbon?

The ring's presence demonstrates that it is unsaturated. Keep in mind that the general formula for aliphatic hydrocarbons, CnH2n+2, serves as the foundation for its saturation. A chemical is unsaturated if it does not meet this requirement.

Example:

Hexane (C6H14)

C = 6; H = 14 = 2(6) + 2

resulting in hexane becoming saturated.

Cyclohexane(C6H12)

C = 6 and H = 12 do not equal 14 (x)!

cyclohexane is an unsaturated molecule as a result.

Cycloalkanes have the general formula C2H2n as well.

Hence, the given statement is false.

Learn more about the hydrocarbons here,

brainly.com/question/17578846

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A mixture of methanol and methyl acetate contains 15.0 wt% methanol.

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The mixture flow rate in lbm/h = 117.65 lbm/h

<h3>Further explanation</h3>

Given

15.0 wt% methanol

The flow rate of the methyl acetate :100 lbm/h

Required

the mixture flow rate in lbm/h

Solution

mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :

$\tt 15\%\times 200~kg=30~kg\\\\mol=\dfrac{mass}{MW}=\dfrac{30~kg}{32~kg/kmol}=0.9375~kmol$

mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :

$\tt 85\%\times 200=170~kg\\\\mol=\dfrac{170}{74}=2.297~kmol$

Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.

1 kg mixture = 0.85 .methyl acetate

So flow rate for mixture :

$\tt \dfrac{1~kg~mixture}{0.85~methyl~acetat}\times 100~lbm/h=117.65~lbm/h$

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Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

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Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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