Explain why has a higher boiling point than NH3

Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su

Ratling [72]

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

1 C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

C₅H₁₂ is the only reactant that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
CO₂ is the only product that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + ? H₂O

Similarly, C₅H₁₂ is the only reactant that contains H atoms. H₂O is the only product that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + 6 H₂O

Both CO₂ and H₂O are products that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the product side corresponds to 8 O₂ molecules on the reactant side.

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", GCSE Chemistry (Single Science), BBC Bitesize.

5 0

3 years ago

Rasheed calculates that his chemical reaction should produce 4 moles of product, but when he does the experiment, he gets only 3

Brut [27]

Answer:

The answer to your question is 75%

Explanation:

Data

Theoretical production = 4 moles

Experimental production = 3 moles

Percent yield = ?

Formula

$Percent yield = \frac{Experimental production}{Theoretical production} x 100$

Substitution

$Percent yield = \frac{3}{4} x 100$

Result

Percent yield = 75 %

8 0

4 years ago

Compare What do infrared technologies, fiber optic

LuckyWell [14K]

Answer:

ptic fiber communication and satellite communication are the leading technologies which are revolutionizing the world of telecommunications. Both technologies have their advantages and limitations which make them suitable for certain type of applications. This article will provide an overview of optic fiber and satellite communication technologies and present a comparison of the features and related issues.

Optic Fiber Communication

Optic Fiber communication transmits information by sending pulses of light (using laser) through an optic fiber. The low signal loss in optic fibers and high data rate of transmission systems, allow signals with high data rates (exceeding several Gbps) to travel over long distances (more than 100 km) without a need of repeater or amplifier. Moreover, using wavelength division multiplexing (WDM) allows a single fiber to carry multiple signals (upto 10 different signals) of multi-Gbps transmissions. Optic Fiber communication offers extremely high bandwidth, immunity to electromagnetic interference, non-existent delays and immunity from interception by external means. In the 1980s and 1990s, the continents were linked together using undersea optic fiber bringing about a paradigm shift in the global telecommunications.

These advancements in optic fiber communication has resulted in decrease of satellite communications for several types of communications. For instance, transmission between fixed locations or point-to-point communications, where large bandwidths are required (such as transoceanic telephone systems) are made through optic fiber instead of using satellite communication. Optic Fiber communication is also used to transmit telephone signals, Internet communication, LAN (Gigabit LAN) and cable television signals.

Satellite Communication

Satellite communications use artificial satellites as relays between a transmitter and a receiver at different locations on Earth. Satellite systems allow users to bypass typical carrier offices and to broadcast information to multiple locations. Communications satellites are used for radio, TV, telephone, Internet, military and other applications. There are more than 2,000 satellites around Earth’s orbit, being used for communication by both government and private organizations.

Communication Satellites are LOS (line-of-sight) microwave systems with a repeater. These satellites rotate around the earth with the speed of earth and are known as geostationary satellites. The limitations of antenna size also limits focusing capability making the coverage for a single satellite transmitter very large. This makes satellite communication ideal for TV and radio services as the signal has to flow from a single point to many points in a single direction. The large distance of satellites from the earth (about 22,300 miles) results in delays which adversely effects two-way communication like mobile conversations. Low earth orbit satellites can be used for two-way mobile communication because less power is required to reach those satellites.

Explanation:

6 0

3 years ago

Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r

Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)

1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑ H bonds broken - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .

8 0

3 years ago

Will mark BRAINLIEST

likoan [24]

Answer:

A and D

Explanation:

4 0

3 years ago