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1 year ago

During complete combustion, natural gas combines with oxygen to produce _____.

Chemistry

1 answer:

MA_775_DIABLO [31]1 year ago

3 0

Answer:

During complete combustion, natural gas combines with oxygen to produce carbon dioxide and water vapor.

Explanation:

A fuel undergoes a complete combustion process when it swiftly combines with oxygen (O2), which then forms carbon dioxide (CO2) and water (H2O).

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What compound is this?

Nezavi [6.7K]

Answer:

Carbon Tetrachloride

Explanation:

1 Carbon atom, 4 chlorine atoms (hence "tetra" prefix)

7 0

3 years ago

Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

$r=[A]^{1}[B]^{2}$ .............(3)

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

new rate r' =

r' = 1/4 r

7 0

3 years ago

Why sodium chloride is a poor conductor of electricity in solid state?

aev [14]

Answer:

Because the electrons in this ionic compound arent free to move and so cannot carry charge. For an iconic compound to conduct electricity it must be a liquid, either in a molten form or dissolved in water.

Explanation:

Is this clear?

5 0

2 years ago

A solution of acetic acid, CH3CO₂H(aq), is at equilibrium. How would the

s344n2d4d5 [400]

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation. (Option C)

<h3>How do systems at equilibrium respond to perturbation?</h3>

When a system at equilibrium suffers a perturbation, it shifts its equilibrium position to counteract such perturbation.

Let's consider a solution of acetic acid at equilibrium.

CH₃CO₂H(aq) = CH₃CO₂⁻(aq) + H⁺(aq)

If more acetic acid were added to the solution, the system will shift toward the products to counteract such an increase.

How would the system change if more acetic acid were added to the solution?

A. [H⁺] would decrease and [CH₃CO₂⁻] would increase. NO.

B. [H⁺] and [CH₃CO₂⁻] would decrease. NO.

C. [H⁺] and [CH₃CO₂⁻] would increase. YES. Both products would increase.

D. [H⁺] would increase and [CH₃CO₂⁻] would decrease. NO.

If more acetic acid were added to a solution at equilibrium, [H⁺] and [CH₃CO₂⁻] would increase to counteract the perturbation.

Learn more about equilibrium here: brainly.com/question/2943338

#SPJ1

3 0

2 years ago

g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.

Lady bird [3.3K]

Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

6 0

3 years ago