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Veronika [31]
3 years ago
9

The complete combustion (reaction with oxygen) of liquid octane (c8h18) a component typical of the hydrocarbons in gasoline, pro

duces carbon dioxide gas and water vapor. what is the coefficient of carbon dioxide gas in the balanced equation for the reaction? (balance the equation with the smallest possible whole number coefficients.)
Chemistry
1 answer:
dmitriy555 [2]3 years ago
4 0
<span>During complete combustion, the hydrocarbon reacts with oxygen (O2) to from carbon dioxide (CO2) and water (H2O). For the combustion of one molecule of octane, 8 molecules of CO2 and 9 molecules of H2O must be formed to account for all the atoms of carbon and hydrogen in the octane. This requires 25 atoms of oxygen, or 12.5 molecules of O2: C8H18+12.5*O2=9*H2O+8*CO2 Multiply both sides by 2 to obtain whole number coefficients: 2*C8H18+25*O2=18*H2O+16*CO2 The coefficient of carbon dioxide in the balanced equation is 16.</span>
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In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b
kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

Explanation:

Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

The reaction is first order in [Co(NH_3)5Br]^{2+}:

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=0.100 M

a) Final concentration of [Co(NH_3)5Br]^{2+} after 19.0 hours= [A]

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

6 0
3 years ago
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

3 0
3 years ago
has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a
Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

               A + B \rightarrow C

Initial :     0.3   0.4          0

Change:  -x       -x           x

Equilbm: (0.3 - x)  (0.4 - x)  x  

We know that, relation between standard free energy and equilibrium constant is as follows.

      \Delta G = -RT ln K

Putting the given values into the above formula as follows.

      \Delta G = -RT ln K

      -3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})

                x = 0.1417

Hence, at equilibrium

  •  [A] = 0.3 - 0.1417

       = 0.1583 M

  •  [B] = 0.4 - 0.1417

       = 0.2583 M

  •  [C] = 0.1417 M
5 0
3 years ago
Multiple Choice Questions 1. Which of the following statement best describes an atom? A. Protons and electrons are grouped toget
Arturiano [62]

Answer:

C is Correct

Explanation:

Protons and Neutrons make up the Nucleus while Electrons surround the nucleus in circular orbits with continuous rotation.

5 0
2 years ago
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Complete the paragraph below. Do it using your Science journal or notebook
svetlana [45]

Answer:

SUSPENSION IS A KIND OF MIXTURE where particlea are visible to the naked eyes AS IT settled AT THE BOTTOM WHEN LEFT UNDISTURBED.

3 0
3 years ago
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