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1 year ago

Which of the following statements is not true of the combustion of an organic compound?

Chemistry

1 answer:

Novay_Z [31]1 year ago

7 0

The following statements are not true of the combustion of an organic compound is 

C. it can also be described as a synthesis reaction 

<h3>Further explanation</h3>

Organic compounds have a characteristic that there is a chain of carbon atoms, while in inorganic does not have a chain of carbon atoms

The reaction of a substance with oxygen is called a combustion reaction

Perfect combustion of carbon compounds will get CO2 gas, whereas if it is not perfect it will produce CO gas

Identification of carbon compounds can usually be done by flowing it into lime water which produces turbid water because of the reaction of CO2 with Ca(OH)₂ to form CaCO₃

Burning organic samples will convert C atoms to CO₂ and H to H₂O

We see the answer choices

1. Oxygen is included in the reactants in the combustion process

So the answer is correct

2. Energy is released during combustion because the reaction is exothermic or releases heat with a negative enthalpy of combustion

So the answer is correct

3. A decomposition reaction is the decomposition reaction of a compound into its constituent elements. In combustion reactions, occur decomposition, not a formation reaction

So the answer is wrong

4. CO₂ is the result of combustion,

So the answer is correct

<h3>Learn more</h3>

properties of hydrocarbons brainly.com/question/4619751

<h3>Answer details </h3>

Grade: Senior High School

Subject: Chemistry

Chapter: Hydrocarbons

Keywords: combustion, organic, inorganic, a carbon chain

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What volume is equivalent to 0.0015 m3?

vaieri [72.5K]

Answer : Option 'D' is correct.

$0.0015m^{3}$ = $1.5\times 10^{6}mm^{3}$

Explanation 1) : 1 m = 100 cm

$1 m^{3}= 1000000 cm^{3}$ or $1 m^{3}= 10^{6} cm^{3}$

$(0.0015m^{3})\times \frac{(1000000cm^{3})}{(1m^{3})}$

$(0.0015)\times \frac{(1000000cm^{3})}{(1)}$ = $1500cm^{3}$

$0.0015m^{3}$ = $1500cm^{3}$ = $1.5\times 10^{3}cm^{3}$

Explanation 2) : 1 m = 1000 mm

$1m^{3}= 1000000000mm^{3} = 10^{9}mm^{3}$

$(0.0015m^{3})\times \frac{(10^{9}mm^{3})}{(1m^{3})}$

$(0.0015)\times \frac{(10^{9}mm^{3})}{(1)}$ = $1500000 mm^{3}$

$0.0015m^{3}$ = $1500000 mm^{3}$ = $1.5\times 10^{6}mm^{3}$

Option 'D' is right.

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2 years ago

Hydrogen bonds are approximately _____% of the bond strength of covalent c-c or c-h bonds.

Lelu [443]

Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.

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1 year ago

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

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1 year ago

A solution of sodium acetate (ch3coona) in water is weakly basic. a. True b. False

guajiro [1.7K]

Hello!

The statement that a solution of sodium acetate (CH₃COONa) is weakly basic is true:

Sodium acetate is the conjugate base of Acetic Acid. When sodium acetate is dissolved in water, it follows the equation that is shown below:

CH₃COONa(s) → CH₃COO⁻(aq) + Na⁺(aq)

Now the Acetate (CH₃COO⁻) ion, has an equilibrium in water to produce hydroxyl (OH⁻) ions and (Acetic Acid CH₃COOH)

CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻

This is a weak equilibrium, and the hydroxyl ions cause the solution to be weakly basic.

Have a nice day!

7 0

2 years ago

(a) What are the possible values of l for n = 4? (Enter your answers as a comma-separated list.)

vodka [1.7K]

Answer:

(a) 0,1,2,3 (b) -3,-2,-1,0,1,2,3 (c) 6 (d) 5

Explanation:

(a) for the principal quantum number 'n', the possible values of I = 0 to n-1. Thus, if the principal quantum number 'n' =4, I = 0,1,2,3.

(b) for a given number of 'I', the possible values of ml = -I to +I. Therefore, if I =3, then ml = -3,-2,-1,0,1,2,3

(c) 'I' which is the orbital angular momentum quantum number usually has values from 0,1,2,⋯,n−1. Therefore, for n greater than or 6, t would be greater than or equal to 5. Thus, the smallest possible value of n for which I can be 6 is 6.

(d) In a 3-dimensional figure, If the z-component of the orbital angular momentum Lz for which I=5 is measured, The possible outcomes will be:

mħ = -5ħ, -4ħ, -3ħ, -2ħ, -1ħ, 0, 1ħ, 2ħ, 3ħ, 4ħ, 5ħ.

Thus, the smallest possible l that can have a z component of 5ℏ is 5.

4 0

2 years ago