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babunello [35]
3 years ago
11

How many alcohols and ethers share the molecular formula c2h6o?

Chemistry
2 answers:
Katen [24]3 years ago
8 0

Answer:

Alcohol :CH₃-CH₂-OH (ethanol)

Ether: H₃C-O-CH₃  (dimethyl ether)

Explanation:

Alcohols are compounds containing the hydroxyl or the -OH functional group i.e. R-OH where R is an alkyl or aryl group

Ethers are compounds with a general structure, R-O-R' where R and R' are once again either alkyl or aryl groups.

The given compound has a molecular formula: C₂H₆O

Structure with 'alcohol' -OH functional group

CH₃-CH₂-OH ------- ethanol

Structure with ether functionality:

H₃C-O-CH₃     ---------dimethyl ether

Nadya [2.5K]3 years ago
4 0

Answer is: there are two compounds with molecular formula C₂H₆O, one alcohol and one ethar.

Alcohol with molecular formula C₂H₆O is ethanol or ethyl alcohol (C₂H₅OH).

Ethanol is a volatile, flammable, colorless liquid with characteristic odor.

Ether with molecular formula C₂H₆O is dimethyl ether (CH₃OCH₃).

Dimethyl ether is a colorless gas.

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If Kc = 4.0×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 520 K , what is the value of Kp for this reaction at this temperature?
Flauer [41]

Here we have to get the K_{p} of the reaction at 520 K temperature.

The K_{p} of the reaction is 1.705 atm

We know the relation between K_{p} and K_{c} is K_{p}=K_{c}(RT)^{N}, where  K_{p} = The equilibrium constant of the reaction in terms of partial pressure, K_{c}  = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.

Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅

Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole

The given value of  K_{c} is 4.0×10⁻²

The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.

On plugging the values in the equation we get,

K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}

Or, K_{p} = 1.705 atm

Thus, the K_{p} of the reaction is 1.705 atm

7 0
3 years ago
a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?
iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

P_{B}=x_{B}*P_{B}^{o}

P_{T}=x_{T}*P_{T}^{o}

Where:

P_{B} is partial pressure for benzene in the liquid  

x_{B} is benzene molar fraction in the liquid  

P_{B}^{o} vapor pressure for pure benzene.  

The total pressure in the solution is:

P= P_{T}+ P_{B}

And  

1=x_{B}+x_{T}

Working on the equation for total pressure we have:

P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}

Since x_{T}=1-x_{B}

P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}

We know P and both vapor pressures so we can clear x_{B} from the equation.

x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}

x_{B}=\frac{35- 22}{75-22} = 0.24

So  

x_{T}=1-0.24 = 0.76

To get the mole fraction for the vapor we know that in the equilibrium:

P_{B}=y_{B}*P

y_{T}=1-y_{B}

So  

y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}

y_{B}=\frac{0.24*75}{35}=0.51

y_{T}=1-0.51=0.49

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

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