Answers
The answer is; Transverse waves
Fiber optic cables transmit information using monochromatic light pulses. All electromagnetic waves (light included) are transverse waves. This means that the particles move perpendicular to the direction of the wave. This is unlike sound waves that are longitudinal waves (particles move parallel to the direction of the wave).
Final answer:
Fiber optic cables transmit data using light waves, which are essentially electromagnetic waves. The wave characteristics of light, especially total internal reflection, interference, and diffraction, facilitate effective data transmissions through these fibers. Factors like high bandwidth, low signal loss, and reduced crosstalk further contribute to their advantage over traditional cables.
Explanation:
Based on the principles of optics, electromagnetic waves, particularly light waves, are what you could find in a fiber optic cable. Fiber optic cables work by transmitting data as pulses of light through strands of fiber made from glass or plastic. This process utilizes the characteristic phenomenon of total internal reflection. When light rays enter the fiber, they bounce off the walls of the fiber cable, undergoing multiple total internal reflections, which ensures that no light escapes the fiber and all signals are conveyed effectively.
Light's wave characteristics are crucial in enabling this functionality. The wave nature of light helps explain properties such as interference and diffraction, essential for the transmission of data in fiber optic networks. These principles are especially relevant when light interacts with small objects such as the core/cladding of the fiber, a subject area often referred to as wave or physical optics.
Another advantage is the high bandwidth of fiber optics, made possible because lasers can emit light with characteristics that allow far more data transmission than electric signals on a single conductor. Meanwhile, properties like low loss and reduced crosstalk enhance the functional superiority of fiber optic cables over traditional copper cable systems.
Learn more about Fiber Optic Cables here:
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Related Questions
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Answers
Answer:
30.1 g NaCl
Explanation:
Your first conversion is converting grams NaOH to moles of NaOH using its molar mass (39.997 g/mol). Then, use the mole ratio of 1 mol NaCl for every 1 mol NaOH to get to moles of NaCl. Then finally multiply by the molar mass of NaCl (58.44 g/mol) to get grams of NaCl.
20.6 g NaOH • (1 mol NaOH / 39.997 g NaOH) • (1 mol NaCl / 1 mol NaOH) • (58.44 g NaCl / 1 mol NaCl) = 30.1 g NaCl
Answers
Answer:
1a. The balanced equation is given below:
2NO + O2 → 2NO2
The coefficients are 2, 1, 2
1b. 755.32g of NO2
2a. The balanced equation is given below:
2C6H6 + 15O2 → 12CO2 + 6H2O
The coefficients are 2, 15, 12, 6
2b. 126.25g of CO2
Explanation:
1a. Step 1:
Equation for the reaction. This is given below:
NO + O2 → NO2
1a. Step 2:
Balancing the equation. This is illustrated below:
NO + O2 → NO2
There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:
2NO + O2 → 2NO2
The equation is balanced.
The coefficients are 2, 1, 2
1b. Step 1:
Determination of the limiting reactant. This is illustrated below:
2NO + O2 → 2NO2
From the balanced equation above, 2 moles of NO required 1 mole of O2.
Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.
From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.
1b. Step 2:
Determination of the maximum amount of NO2 produced. This is illustrated below:
2NO + O2 → 2NO2
From the balanced equation above, 2 moles of NO produced 2 moles of NO2.
Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.
1b. Step 3:
Conversion of 16.42 moles of NO2 to grams. This is illustrated below:
Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol
Mole of NO2 = 16.42 moles
Mass of NO2 =?
Mass = number of mole x molar Mass
Mass of NO2 = 16.42 x 46
Mass of NO2 = 755.32g
Therefore, the maximum amount of NO2 produced is 755.32g
2a. Step 1:
The equation for the reaction.
C6H6 + O2 → CO2 + H2O
2a. Step 2:
Balancing the equation:
C6H6 + O2 → CO2 + H2O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:
C6H6 + O2 → 6CO2 + H2O
There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:
C6H6 + O2 → 6CO2 + 3H2O
There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:
C6H6 + 15/2O2 → 6CO2 + 3H2O
Multiply through by 2 to clear the fraction.
2C6H6 + 15O2 → 12CO2 + 6H2O
Now, the equation is balanced.
The coefficients are 2, 15, 12, 6
2b. Step 1:
Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:
2C6H6 + 15O2 → 12CO2 + 6H2O
Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol
Mass of C6H6 from the balanced equation = 2 x 78 = 156g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 15 x 32 = 480g
2b. Step 2:
Determination of the limiting reactant. This is illustrated below:
From the balanced equation above,
156g of C6H6 required 480g of O2.
Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.
From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.
2b. Step 3:
Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb
2C6H6 + 15O2 → 12CO2 + 6H2O
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 12 x 44 = 528g
2b. Step 4:
Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:
From the balanced equation above,
156g of C6H6 produced 528g of CO2.
Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2
b. Negative
c. Zero
d. Impossible to determine
Answers
Answer: The standard free energy change for a reaction in an electrolytic cell is always positive.
Explanation:
Electrolytic cells use electric currents to drive a non-spontaneous reaction forward.
Relation of standard free energy change and emf of cell
where,
= standard free energy change
n= no of electrons gained or lost
F= faraday's constant
= standard emf
= standard emf = -ve , for non spontaneous reaction
Thus
Thus standard free energy change for a reaction in an electrolytic cell is always positive.
Answers
Answer:
1. 2Al + 3I2 —> Al2I6
2. 0.555mol of I2
Explanation:
1. Al + I2 —> Al2I6
Observing the above equation, there are 2 atoms of Al on the right side and 1 on the left side. To balance it, put 2 in front of Al as shown below:
2Al + I2 —>Al2I6
Also, there are 6 atoms of I on the right side and 2 on the left side. To balance it, put 3 in front I2 as shown below:
2Al + 3I2 —>Al2I6
2. Molar Mass of Al = 27g/mol
Mass of Al = 10g
n = Mass /Molar Mass
n = 10/27 = 0.37mol
From the equation,
2moles of Al reacted with 3 moles of I2.
Therefore, 0.37mol of Al will react with = (0.37 x 3)/2 = 0.555mol of I2
The balanced chemical equation is:
2 Al(s) + 3 I₂(s) → Al₂I₆(s)
0.557 moles of iodine react with 10.0 g of aluminum.
Let's consider the following unbalanced equation.
Al(s) + I₂(s) → Al₂I₆(s)
We will balance it using the trial and error method. We can get the balanced equation by multiplying Al by 2 (balance Al atoms) and I₂ by 3 (balance I atoms).
2 Al(s) + 3 I₂(s) → Al₂I₆(s)
The molar mass of Al is 26.98 g/mol. The moles corresponding to 10.0 g of Al are:
The molar ratio of Al to I₂ is 2:3. The moles of I₂ that react with 0.371 moles of Al are:
The balanced chemical equation is:
2 Al(s) + 3 I₂(s) → Al₂I₆(s)
0.557 moles of iodine react with 10.0 g of aluminum.
You can learn more about stoichiometry here: brainly.com/question/9743981
Answers
Among all three PCl₃, NO₃⁻ , I₃⁻, H₂Se only I₃⁻ will involve participation of d-orbitals in hybridization as bonding in I₃⁻ will include, s, p and d orbital as shown in the image attached. there will be sp³d hybridization, there will be presence of three lone pairs and 2 bonds as
I⁻ has 8 valence electrons and 2 neighbours atoms which will need one electron each to satisfy their valency.
so the number of electrons on central atom will be:
8-1-1=6
That 6 electrons will make 6/2 =3 lone pairs.
b. catalytic turnover more than substrate binding.
c. substrate binding, but not catalytic turnover.
d. catalytic turnover, but not substrate binding.
Answers
Answer:
Explanation:
The various substitutions affect catalytic turnover more than substrate binding.