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4 years ago

Draw the lewis structure for the molecule ch2chch3. how many sigma and pi bonds does it 30) contain

Chemistry

1 answer:

juin [17]4 years ago

8 0

Answer:
1-Propene has eight sigma bonds and one pi bond.

Explanation:
The lewis structure of 1-Propene is shown below. All sigma bonds are highlighted by blue color and the pi bond is highlighted by red color.
Carbon having four unpaired electrons in its excited state can form four covalent (sigma bonds) bonds. In given structure the orbitals containing unpaired electron in sp³ hybridized C undergo head to head overlap with three hydrogen atoms having single unpaired electron and orbital of other sp² hybridized carbon in same fashion. So, four sigma bonds are formed by sp³ hybridized carbon atom.
The doubly bonded carbon atoms also form sigma bond but two unpaired electrons of these two carbons undergo the formation of pi bond. This pi bond forms when the orbitals overlap along the plane perpendicular to the existing sigma bond resulting in the formation of single pi bond.

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In laboratory experiment, a NOVDEC Student was

Ede4ka [16]

Answer:

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

The volume of the glucose solution to be prepared = 500 $cm^3$

Molarity of the glucose solution to be prepared = 1 M

i. Molar mass of glucose ( $C_1_2H_6O_6$ ) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii. mole = molarity x volume. Hence;

amount (in moles) of the glucose solution to be prepared

= 1 x 500/1000 = 0.5 mole

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3 years ago

How do you label the delta E, on an energy diagram

vladimir1956 [14]

Delta energy on labelled diagram is attached below

4 0

3 years ago

In a basic solution

Gnom [1K]

Answer:

Explanation:

OH is generally basic, while H is generally acidic so the greater the OH concentration, the more basic the solution will be.

4 0

4 years ago

Solution a is a 1.0 m solution with a nonionic solute and water as the solvent. solution b is a 1.0 m solution with the same non

solniwko [45]

The solution of ethanol will have the greatest increase in boiling point.
The formula for boiling point elevation is:
Î”Tb = Kb Â· bB
where
Î”Tb = boiling point elevation
Kb = ebullioscopic constant for the solvent
bB = molarity of the solution
Since in the solute is nonionic, we don't have to worry about the molecules of the solute breaking up into multiple ions, thereby increasing the effective molarity of the solution. So which ever solvent has the highest ebullioscopic constant, will have the greatest increase in boiling point. This constant can be calculated by the equation:
Kb = RTb^2M/Î”Hv
where
R = Ideal gas constant
Tb = boiling point of pure solvent
M = Molar mass of solvent
Î”Hv = heat of vaporization per mole of solvent
For our purposes, we can ignore the idea gas constant, and instead look at only the boiling point, molar mass, and heat of vaporization. Then calculate Tb^2M/Î”Hv So let's do so:
(Note: Not bothering to be precise in molar mass. If the end result is close, then I'll bother. Otherwise, just using nice round numbers).
Water
Boiling point: 373.15 K
Molar mass: 18 g/mol
heat of vaporization: 40660 J/mol
Tb^2M/Î”Hv: 61.64
Ethanol
Boiling point: 351.52 K
Molar mass: 46 g/mol
heat of vaporization: 38600 J/mol
Tb^2M/Î”Hv: 147.26
The value of Tb^2M/Î”Hv is significantly greater for ethanol than it is for water (by more than 2 to 1), so it will have the greatest increase in boiling point.

4 0

3 years ago

Read 2 more answers

A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of

Goryan [66]

Answer:

C₂ H₄ O

Explanation:

1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)

atomic mass of C: 12.0107 g/mol

molar mass of CO₂: 44.01 g/mol

Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂

Solve for x:

x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C

2) Mass of hydrogen (H) in 0.930 g of water (H₂O)

atomic mass of H: 1.00784 g/mol

molar mass of H₂O: 18.01528 g/mol

proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O

Solve for x:

x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H

3) Mass of oxygen (O) in 1.0857 g of pure sample

Mass of O = mass of pure sample - mass of C - mass of H

Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

4) Mole calculations

Divide the mass in grams of each element by its atomic mass:

C: 0.59767 g / 12.0107 g/mol = 0.04976 mol

H: 0.10406 g / 1.00784 g/mol = 0.10325 mol

O: 0.3840 g / 15.999 g/mol = 0.02400 mol

5) Divide every amount by the smallest value (to find the mole ratios)

C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2

H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4

O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

C₂ H₄ O ← answer

3 0

3 years ago