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2 years ago

What is the empirical formula of propene CH3 CH2 CH2

Chemistry

1 answer:

kiruha [24]2 years ago

6 0

Answer:

CH

Explanation:

Molecular formula of propene : C₃H₆

Take the HCF of carbon and hydrogen atoms :

3 = 3
6 = 2 x 3

Then, we can write the formula as :

3CH
This means there are 3 moles present

Empirical Formula :

Molecular Formula / No. of moles
C₃H₆ / 3
CH

The empirical formula of propene is CH

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What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigonal planar arrang

Ksivusya [100]

Answer:

The required hybridization for a central atom that have:

a tetrahedral arrangement of electron pairs is $sp^3$

a trigonal planar arrangement of electron pairs is $sp^2$

a linear arrangement of electron pairs is $sp$

Explanation:

From the given information:

The required hybridization for a central atom that have:

a tetrahedral arrangement of electron pairs is $sp^3$

a trigonal planar arrangement of electron pairs is $sp^2$

a linear arrangement of electron pairs is $sp$

Hybridization is the mixing and blending of two or more pure atomic orbitals ( s, p and d) to forma two or more hybrid atomic orbitals that are identical in shape and energy e. sp ,sp² , sp³, sp³d hybrid orbitals.

Examples:

Tetrahedral

In CH₄ , carbon C is the central atom.

A 2s electron is excited from the ground state of boron 1s²2s²2p² to one of the empty orbitals to 2p to give the excited state 1s²2s²2p³.

In the excited state of carbon, the 2-s orbital can be mixed with the 2p orbitals in three ways : sp³, sp² and sp hybridization. For the formation of four sp³ hybrid orbitals, the 2s orbital are mixed with all the three p orbitals. The four sp³ hybrid orbitals are tetrahedrally arranged with a bond angle of 109.5⁰

Trigonal Planar

In BF₃ , Boron B is the central atom

A 2s electron is excited from the ground state of boron 1s²2s²2p¹ to one of the empty orbitals to 2p. The 2s orbital is then mixed with two orbitals of 2p to form three sp² hybrid orbitals that are trigonal planar arranged in the plane in order to minimize repulsion. They bonds between them form an equal strength and with bond angles of 120⁰

Linear arrangement:

In BeCl₂, Be is the central atom

To provide the unpaired electrons for covalent bonds a 2s electron is excited to a 2p orbital. Thereafter, the two atomic orbitals hybridize to give two identical orbitals called sp hybrid orbitals. The sigma bond for,ed are equal in bond lengths and form a bond angle of 180°

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Water molecules have a polarity, which allows them to be electrically attracted to other water molecules and other polar molecul

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Water molecules have a polarity, which allows them to be electrically attracted to other water molecules and other polar molecules by Hydrogen bonds.

Hydrogen Bond

Between a hydrogen (H) atom that is covalently bonded to a more electronegative "donor" atom or group and another electronegative atom bearing a lone pair of electrons—the hydrogen bond acceptor—a hydrogen bond (or H-bond) forms (Ac). The second-row elements fluorine, oxygen, and nitrogen are the most common donor and acceptor atoms (F).

It may take place either intramolecularly or intermolecularly (between different molecules) (occurring among parts of the same molecule). The energy of a hydrogen bond can range from 1 to 40 kcal/mol and is influenced by the shape, surroundings, and nature of the particular donor and acceptor atoms.

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2 years ago

Read 2 more answers

A. The reactant concentration in a zero-order reaction was 8.00×10−2 M after 155s and 3.00×10−2 M after 355s . What is the rate

irga5000 [103]

Answer:

A) The rate constant is 2.50 × 10⁻⁴ M/s.

B) The initial concentration of the reactant is 11.9 × 10⁻² M.

C) The rate constant is 0.0525 s⁻¹

D) The rate constant is 0.0294 M⁻¹ s⁻¹

Explanation:

Hi there!

A) The equation for a zero-order reaction is the following:

[A] = -kt + [A₀]

Where:

[A] = concentrationo f reactant A at time t.

[A₀] = initial concentration of reactant A.

t = time.

k = rate constant.

We know that at t = 155 s, [A] = 8.00 × 10⁻² M and at t = 355 s [A] = 3.00 × 10⁻² M. Then:

8.00 × 10⁻² M = -k (155 s) + [A₀]

3.00 × 10⁻² M = -k (355 s) + [A₀]

We have a system of 2 equations with 2 unknowns, let´s solve it!

Let´s solve the first equation for [A₀]:

8.00 × 10⁻² M = -k (155 s) + [A₀]

8.00 × 10⁻² M + 155 s · k = [A₀]

Replacing [A₀] in the second equation:

3.00 × 10⁻² M = -k (355 s) + [A₀]

3.00 × 10⁻² M = -k (355 s) + 8.00 × 10⁻² M + 155 s · k

3.00 × 10⁻² M - 8.00 × 10⁻² M = -355 s · k + 155 s · k

-5.00 × 10⁻² M = -200 s · k

-5.00 × 10⁻² M/ -200 s = k

k = 2.50 × 10⁻⁴ M/s

The rate constant is 2.50 × 10⁻⁴ M/s

B) The initial reactant conentration will be:

8.00 × 10⁻² M + 155 s · k = [A₀]

8.00 × 10⁻² M + 155 s · 2.50 × 10⁻⁴ M/s = [A₀]

[A₀] = 11.9 × 10⁻² M

The initial concentration of the reactant is 11.9 × 10⁻² M

C) In this case, the equation is the following:

ln[A] = -kt + ln([A₀])

Then:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

Let´s solve the first equation for ln([A₀]) and replace it in the second equation:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(7.60 × 10⁻² M) + 35.0 s · k = ln([A₀]

Replacing ln([A₀]) in the second equation:

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln(7.60 × 10⁻² M) + 35.0 s · k

ln(5.50 × 10⁻³ M) - ln(7.60 × 10⁻² M) = -85.0 s · k + 35.0 s · k

ln(5.50 × 10⁻³ M) - ln(7.60 × 10⁻² M) = -50.0 s · k

(ln(5.50 × 10⁻³ M) - ln(7.60 × 10⁻² M)) / -50.0 s = k

k = 0.0525 s⁻¹

The rate constant is 0.0525 s⁻¹

D) In a second order reaction, the equation is as follows:

1/[A] = 1/[A₀] + kt

Then, we have the following system of equations:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

Let´s solve the first equation for 1/[A₀]:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/ 0.510 M - 205 s · k = 1/[A₀]

Now let´s replace 1/[A₀] in the second equation:

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

1/5.10 × 10⁻² M = 1/ 0.510 M - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = 600 s · k

(1/5.10 × 10⁻² M - 1/ 0.510 M)/ 600 s = k

k = 0.0294 M⁻¹ s⁻¹

The rate constant is 0.0294 M⁻¹ s⁻¹

8 0

3 years ago

Given the equation:

Citrus2011 [14]

Answer: a) $Fe$ is the limiting reagent

b) 3.59 g

c) 11.6g

Explanation:

$4Al_2O_3+9Fe\rightarrow 3Fe_3O_4+8Al$

To calculate the moles :

$\text{Moles of} Al_2O_3=\frac{27.5g}{102g/mol}=0.27moles$

$\text{Moles of} Fe=\frac{8.4g}{56g/mol}=0.15moles$

According to stoichiometry :

a) 9 moles of $Fe$ require= 4 moles of $Al_2O_3$

Thus 0.15 moles of $Fe$ will require= $\frac{4}{9}\times 0.15=0.067moles$ of $Al$

Thus $Fe$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

b) As 9 moles of $Fe$ give = 8 moles of $Al$

Thus 0.15 moles of $Fe$ give = $\frac{8}{9}\times 0.15=0.133moles$ of $Al$

Mass of $Al=moles\times {\text {Molar mass}}=0.133moles\times 27g/mol=3.59g$

c) As 9 moles of $Fe$ give = 3 moles of $Fe_3O_4$

Thus 0.15 moles of $Fe$ give = $\frac{3}{9}\times 0.15=0.05moles$ of $Fe_3O_4$

Mass of $Fe_3O_4=moles\times {\text {Molar mass}}=0..05moles\times 231.5g/mol=11.6g$

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3 years ago

Do you think lava or water formed the Chanel on Mars? And why? I’ll give brainliest

Kamila [148]

Answer:

A channel in an ancient Martian "river bed" was not carved out by liquid water but built by molten lava

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3 years ago

What is the empirical formula of propene CH3 CH2 CH2​

What is the empirical formula of propene CH3 CH2 CH2