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3 years ago

How do you write a balanced equation for the combustion ofbenzene?

Chemistry

1 answer:

forsale [732]3 years ago

4 0

Answer: The balanced chemical equation for the combustion of benzene is:

$2C_6H_6+15O_2\rightarrow 12CO_2+6H_2O$

Explanation:

Combustion reactions : It is defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water as a product.

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

The balanced chemical equation for the combustion of benzene will be:

$2C_6H_6+15O_2\rightarrow 12CO_2+6H_2O$

By Stoichiometry of the reaction we can say that,

2 mole of benzene reacts with 15 moles of oxygen gas to produce 12 moles of carbon dioxide and 6 moles of water molecule.

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘c is 2000 m/s. Note that 1. 0 mol of diatomic hydrogen at

Leno4ka [110]

The rms speed will be 500 m/s

<h3>What is Root mean square speed ?</h3>

Root mean square speed (Vrms) is defined as the square root of the mean of the square of speeds of all molecules.

Root mean square speed (vrms) Root mean square speed (vrms) is defined as the square root of the mean of the square of speeds of all molecules

It is given that

Speed of a diatomic hydrogen molecule,2000 m/s

Mol of diatomic hydrogen,1.0

Temperature,50°C

The rms speed of diatomic molecule will be:

√(3KT)/( m)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

(3/2)KT = (1/2) mv²

v = √(3KT)/m

FOR H₂: v = √(3KT)/m = 2000 m/s

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

Therefore the rms (root-mean-square) speed of a oxygen molecule at 50∘c is 500m/s.

To know more about Root mean square speed

brainly.com/question/7213287

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3 0

2 years ago

Match each scientist to their discovery regarding the atom

neonofarm [45]

Answer:

Thomson--atoms cotain electron

Ernest Rutherford--atoms have a positive nucleus

R.A Millikan--electrons have Q=-1

Dalton--atoms are indivisible

3 0

2 years ago

Read 2 more answers

Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to

valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

Explanation:

Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of H₂SO₄.

Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.

To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is = 5.451 X 10³ kg.

5 0

3 years ago

Using the method of initial rates for the reaction a → b, if the initial concentration of a is doubled and the rate of reaction

zmey [24]

Answer is: reaction is second order with respect to a.
This second order reaction is proportional to the square of the concentration of reactant a.
rate of reaction = k[a]².
k is second order rate constant and have unit M⁻¹·s⁻¹.
Integrated rate law for this reaction: 1/[a]=1/[a]₀ + kt.
t is time in seconds..

4 0

2 years ago

Consider the balanced equation for the following reaction:

Bad White [126]

Answer: The theoretical yield of the lithium chlorate is 1054.67 grams

Explanation:

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

$9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g$

To calculate the theoretical yield of lithium chlorate, we use the equation:

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

$81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g$

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0

2 years ago