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4 years ago

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The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

1.95 mg of ethyl butyrate produces 4.42 mg of CO2 and 1.81 mg of H2O. What is the empirical formula of the compound?

1 answer:

Mnenie [13.5K]4 years ago

3 0

Answer:

Empirical formula is C₃H₃O

Explanation:

Given data:

Mass of ethyl butyrate = 1.95 mg

Mass of CO₂ = 4.42 mg

Mass of water = 1.81 mg

Empirical formula = ?

Solution:

Percentage of C = 4.42/1.95 × 12/44×100

= 2.27× 0.273 ×100

= 62

percentage of H = 1.81/1.95 × 2/18 × 100

= 0.93 × 0.11 ×100

= 10.23

percentage of oxygen = 100 - (72.23)

= 27.77

number of gram atoms of C = 62/12 =5.17

number of gram atoms of H = 10.23 / 2 =5.115

number of gram atoms of O = 27.77 / 16 =1.74

Atomic ratio:

C : H : O

5.17/1.74 : 5.115/1.74 : 1.74/1.74

3 : 3 : 1

Empirical formula is C₃H₃O

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A sample of O2(g) is placed in an otherwise empty, rigid container at 4224 K at an initial pressure of 4.97 atm, where it decomp

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Answer:

The value of $K_p$ at 4224 K is 314.23.

Explanation:

$O_2(g)\rightleftharpoons 2O(g)$

Initially

4.97 atm 0

At equilibrium

4.97 - p 2p

At initial stage, the partial pressure of oxygen gas = =4.97 atm

At equilibrium, the partial pressure of oxygen gas = $p_{O_2}=0.28 atm$

So, 4.97 - p = 0.28 atm

p = 4.69 atm

At equilibrium, the partial pressure of O gas = $p_{O}=2p=2\times 4.69 atm=9.38 atm$

The expression of $K_p$ is given as :

$K_p=\frac{(p_{O})^2}{(p_{O_2})}$

$K_p=\frac{(9.38 atm)^2}{0.28 atm}=314.23$

The value of $K_p$ at 4224 K is 314.23.

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3 years ago

The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th

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Answer:

The coefficient of thermal expansion α is

$\alpha = \frac{1}{T}$

The coefficient of compressibility

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

So

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

Explanation:

From the question we are told that

The coefficient of thermal expansion is $\alpha = \frac{1}{V} * (\frac{\delta V}{ \delta P}) P$

The coefficient of compressibility is $\beta = - (\frac{1}{V} ) * (\frac{\delta V}{ \delta P} ) T$

Generally the ideal gas is mathematically represented as

$PV = nRT$

=> $V = \frac{nRT}{P} --- (1)$

differentiating both side with respect to T at constant P

$\frac{\delta V}{\delta T } = \frac{ n R }{P}$

substituting the equation above into $\alpha$

$\alpha = \frac{1}{V} * ( \frac{ n R }{P}) P$

$\alpha = \frac{nR}{PV}$

Recall from ideal gas equation $T = \frac{PV}{nR}$

So

$\alpha = \frac{1}{T}$

Now differentiate equation (1) above with respect to P at constant T

$\frac{\delta V}{ \delta P} = -\frac{nRT}{P^2}$

substituting the above equation into equation of $\beta$

$\beta = - (\frac{1}{V} ) * (-\frac{nRT}{P^2} ) T$

$\beta =\frac{ (\frac{n RT}{PV} )}{P}$

Recall from ideal gas equation that

$\frac{PV}{nRT} = 1$

So

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

So

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

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3 years ago

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

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<u>Explanation:</u>

For the given half reactions:

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Reduction half reaction: $Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V$

Net reaction: $Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.28-(-0.76)=0.48V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

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$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}$

Hence, the $K_{eq}$ of the reaction is $1.73\times 10^{16}$

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