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Zepler [3.9K]
2 years ago
4

Natural gas is used as a cooking fuel in many restaurants and homes. The primary chemical components of natural gas are hydrocar

bons known as alkanes. Research and list the primary alkanes found in natural gas, and explain why these compounds excel as fuel sources.
Chemistry
1 answer:
weqwewe [10]2 years ago
7 0
The main component of natural gas is methane (CH4) at 60 to 90% followed by various combination of ethane, propane, and butane whose percentage can vary from 0 to 20% each. For each unit mass of alkanes, the combustion energy (energy released when the fuel reacts with oxygen) released is very high about 13 to 15 kcal/g, which is higher than even those generated by petrol or diesel. So, for heating or other energy generation purpose for household purposes, this source of energy is used.

The equation for combustion of methane is shown below. Upon combustion, carbondioxide and water is produced with simultaneous generation of heat which is the source of energy used for consumption.

CH4 + 2O2 --> CO2+ 2H2O + heat [ For methane, the combustion energy is ~ 6kcal/g]

As the CH2 units are increased in the alkanes, the combustion energy increases, for e.g., ethane has combustion energy of 7 kcal/g and propane has about 12 kcal/g.
You might be interested in
What is the conjugate acid of each of the following? What is the conjugate base of each?
Lilit [14]

Answer:

a. H₂O (conjugate acid) ; b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid) ; c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base) ; d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base) ; f. No conjugate acid either base;  g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a.  OH⁻  +  H⁺  ⇄ H₂O

The hydroxide acts like a Bronsted Lory base, so it can catch a proton. Water will be the conjugate acid.

b. H₂O, is an amphoterus compound. It can act as an acid or a base. If it is a base, the conjugate acid is the H₃O⁺. If it is an acid, the conjugate base is the OH⁻.

c. HCO₃⁻  +  H⁺  ⇄  H₂CO₃

HCO₃⁻  +  H₂O  ⇄ CO₃⁻²  +  H₃O⁺

The bicarbonate is also amphoteric. When it catches the proton, the carbonic acid is the conjugate acid, cause it works as a base.

When the HCO₃⁻ (acid) release the proton, the conjugate base is the carbonate.

d. Ammonia is a weak base, so the conjugate strong acid is the ammonium.

NH₃ + H₂O  ⇄  NH₄⁺  +  OH⁻

e. Another amphoteric compound. The acid sulfate acts an acid and a base.

(like bicarbonate). When it is a base, the conjugate acid is the sulfuric acid, when it is an acid, the conjugate base is the sulfate.

HSO₄⁻  +  H₂O  ⇄  SO₄⁻²  +  H₃O⁺

HSO₄⁻  +  H⁺  ⇄  H₂SO₄

f. H₂O₂ does not recieve H⁺ or OH⁻, and it does not release H⁺. It is a neutral compound and it doesn't act as a base or acid.

g. HS⁻ is amphoterous.

HS⁻  +  H⁺  ⇄  H₂S

HS⁻  +  H₂O  ⇄  S⁻²  +  H₃O⁺

Same case as bicarbonate or acid sulfate.

h. H₅N₂⁺  +  H₂O  ⇄  H₄N₂  + H₃O⁺

Hidrazinium acts an acid, so, the conjugate base will be the hidrazine.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           

3 0
2 years ago
A solution has a hydroxide-ion concentration of 0.0040 M. What is the pOH of the solution? A solution has a pH value of 3.66. Wh
hodyreva [135]

Answer:

1. pOH = 2.4

2. pOH = 10.34

3. pH = 2.14

Explanation:

1. Determination of the pOH.

Concentration of Hydroxide ion, [OH-] = 0.004 M

pOH =?

pOH = - log [OH-]

pOH = - log (0.004)

pOH = 2.4

Therefore, the pOH of the solution is 2.4

2. Determination of the pOH.

pH = 3.66

pOH =?

pH and pOH are related by the following equation:

pH + pOH = 14

With the above formula, we can obtain the pOH of the solution as follow:

pH = 3.66

pOH =?

pH + pOH = 14

3.66 + pOH = 14

Collect like terms

pOH = 14 - 3.66

pOH = 10.34

Therefore, the pOH of the solution is 10.34

3. Determination of the pH.

Molarity of HCl = 0.0072 M

Concentration of Hydrogen ion, [H+] =?

Thus, we can obtain the concentration of Hydrogen ion, [H+] as follow:

HCl(aq) —> H+(aq) + Cl-(aq)

From the balanced equation above,

1 mole of HCl produced 1 mole of H+.

Therefore, 0.0072 M HCl will also produce 0.0072 M H+.

Therefore, the concentration of Hydrogen ion, [H+] in the solution is 0.0072 M.

Finally, we shall determine the pH of the solution as follow:

Concentration of Hydrogen ion, [H+] = 0.0072 M.

pH =?

pH = - log [H+]

pH = - log (0.0072)

pH = 2.14

Therefore, the pH of the solution is 2.14

3 0
2 years ago
How many mL of a 5.0% glucose solution provide 80.0 g of glucose?
Llana [10]
Maybe
A. 400 ml of 5.0% glucose solution
4 0
2 years ago
A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate
fgiga [73]
First, we are using the ideal gas law to get n the number of moles:

PV = nRT

when P is the pressure = 748 mmHg/760 = 0.984 atm

V is the volume = 4 L

R is ideal gas constant = 0.0821

T is the temperature in Kelvin = 300 K

∴ n =  0.984atm*4L/0.0821*300

       = 0.1598 moles

when the concentration = moles * (1000g / mass)

                                         = 0.1598 * (1000g / 58 g )

                                         = 2.755 M

when the freezing point = 5.5 °C

and Kf = - 5.12 °C/m

∴ the freezing point for the solution = 5.5 °C + (Kf*m)

                                                            = 5.5 °C - (5.12°C/m * 2.755m)

                                                            = -8.6 °C

8 0
2 years ago
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
2 years ago
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