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2 years ago

What is the empirical formula of butenedioic acid

Chemistry

2 answers:

icang [17]2 years ago

7 0

Answer:

C4H4O4

Explanation:

Hope this helps!

garik1379 [7]2 years ago

3 0

The formula is C4H4O4

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Chemistry vocabulary and answer

finlep [7]

Answer:

you forgot to attach a picture :)

Explanation:

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3 years ago

Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10

mylen [45]

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

$C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$

Initial: 24.0

At eqllm: 24-x x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}$

We are given:

$K_p=0.040$

Putting values in above expression, we get:

$0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1$

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

$PV=nRT$ .........(1)

To calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ..........(2)

For ethane gas:

We are given:

$P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol$

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

$7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g$

For ethylene gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g$

For hydrogen gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g$

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

$\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100$

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

$\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%$

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

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Flowers usually contain more stamen than pistils. Why do you think this is ?

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3 years ago

What is the empirical formula of butenedioic acid​

What is the empirical formula of butenedioic acid