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joja [24]
2 years ago
8

"The combustion of ethylene proceeds by the reaction C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) When the rate of disappearance of O2 i

s 0.23 M s-1, the rate of disappearance of C2H4 is ________ M s-1."
Chemistry
1 answer:
Julli [10]2 years ago
7 0

Answer:

The rate of  disappearance of C_2H_4 is 0.0766 M/s.

Explanation:

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)

Rate of the reaction = R

R=-\frac{1}{1}\frac{d[C_2H_4]}{dt}=-\frac{1}{3}\frac{d[O_2]}{dt}

[te]R=\frac{1}{2}\frac{d[CO_2]}{dt}=\frac{1}{2}\frac{d[H_2O]}{dt}[/tex]

Rate of disappearance of O_2 =0.23 M/s

Rate of disappearance of O_2=3\times R

0.23 M/s=3\times R

R =0.07666 M/s

Rate of disappearance of C_2H_4=1\times R

\frac{d[C_2H_4]}{dt}=1\times 0.23 m/s=0.0766 M/s

The rate of  disappearance of C_2H_4 is 0.0766 M/s.

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A proton transfer reaction can occur when an aldehyde is placed in strong base, such as an alkoxide ion, producing an alcohol an
Pani-rosa [81]

Hi, you have not provided structure of the aldehyde and alkoxide ion.

Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.

Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.

The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.

After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.

All the structures are shown below.

7 0
2 years ago
Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t
-Dominant- [34]

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒  ABCC = \frac{4RBCC}{\sqrt{3} }

⇒             = \frac{4\times 0.12584}{\sqrt{3}}

⇒             = 0.29062 \ nm

Likewise AFCC as well as RFCC are interconnected by  

√2AFCC = 4RFCC

⇒  AFCC = \frac{4RFCC}{\sqrt{2}}

⇒             = \frac{4\times 0.12894}{\sqrt{2} }

⇒             = 0.36470 \ nm

Now,

The Change in Percent Volume,

= \frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent

= \frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent

= \frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent

= \frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent

= 97.62 \ percent (approximately)

Note: percent = %

7 0
2 years ago
For a pure substance, the liquid and gaseous phases can only coexist for a single value of the pressure at a given temperature.
anastassius [24]

Answer:

No, it is not.

Explanation:

Most solutions do not behave ideally. Designating two volatile  substances as A and B, we can consider the following two cases:

Case 1: If the intermolecular forces between A and B molecules are weaker than  those between A molecules and between B molecules, then there is a greater tendency  for these molecules to leave the solution than in the case of an ideal solution. Consequently,  the vapor pressure of the solution is greater than the sum of the vapor  pressures as predicted by Raoult’s law for the same concentration. This behavior gives  rise to the positive deviation.

Case 2: If A molecules attract B molecules more strongly than they do their own  kind, the vapor pressure of the solution is less than the sum of the vapor pressures as  predicted by Raoult’s law. Here we have a negative deviation.

The benzene/toluene system is an exception, since that solution behaves ideally.

8 0
2 years ago
In an experiment, 0.42 mol of co and 0.42 mol of h2 were placed in a 1.00-l reaction vessel. at equilibrium, there were 0.29 mol
Likurg_2 [28]
To determine the Keq, we need the chemical reaction in the system. In this case it would be:

CO + 2H2 = CH3OH

The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:

          CO           H2             CH3OH
I         .42           .42                    0
C     -0.13      -2(0.13)            0.13
-----------------------------------------------
E =    .29           0.16               0.13

Therefore, 

Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51
4 0
2 years ago
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u
GenaCL600 [577]

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

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First reaction;

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Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

7 0
1 year ago
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