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3 years ago

How many moles of carbon dioxide are produced from the complete combustion of 5.42 moles of ethanol? 2C2H6O +7 O2 ®4 CO2 + 6H2O

Chemistry

1 answer:

Evgen [1.6K]3 years ago

4 0

Answer: 10.84 moles of $CO_2$ will be produced from complete combustion of 5.42 moles of ethanol

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

$2C_2H_6O+7O_2\rightarrow 4CO_2+6H_2O$

According to stoichiometry :

2 moles of ethanol produce = 4 moles of $CO_2$

Thus 5.42 moles of ethanol will produce= $\frac{4}{2}\times 5.42=10.84moles$ of $CO_2$

Thus 10.84 moles of $CO_2$ will be produced from complete combustion of 5.42 moles of ethanol

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3 years ago

Enough of a monoprotic acid is dissolved in water to produce a 1.64 M solution. The pH of the resulting solution is 2.82 . Calcu

Lady bird [3.3K]

Answer:

Ka = 1.39x10⁻⁶

Explanation:

A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

<em>Where Ka is:</em>

Ka = [H⁺] [X⁻] / [HX]

<em>Where [] is the molar concentration in equilibrium of each specie. </em>

The equilibrium is reached when some HX reacts producing H+ and X-, that is:

[HX] = 1.64M - X

[H⁺] = X

[X⁻] = X

As pH is 2.82 = -log [H⁺]:

[H⁺] = 1.51x10⁻³M:

[HX] = 1.64M - 1.51x10⁻³M = 1.638M

[H⁺] = 1.51x10⁻³M

[X⁻] = 1.51x10⁻³M

And Ka is:

Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]

6 0

3 years ago

Determine the equilibrium constant, Keq, at 25°C for the reaction

adelina 88 [10]

Explanation:

The given chemical reaction is:

$2Br^- (aq) + I_2(s) Br_2(l) + 2I^- (aq)$

$E^ocell=oxidation potential of anode + reduction potential of cathode\\$

The relation between Eo cell and Keq is shown below:

$deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell$

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

$E^ocell= (-1.07+0.53)V\\=-0.54V$

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

$ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3$

Answer:

Keq=6.13x10^33

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3 years ago

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alexgriva [62]

When an iron is dipped in Copper Sulphate

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FeSO4 is light green color.

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3 years ago