The correct answer for the question that is being presented above is this one: "<span>0.3."
Here it is how to solve.
M</span><span>olecular mass of Ar = 40
</span><span>Molecular mass of Ne = 20
</span><span>Number of moles of Ar = 9.59/40 = 0.239
</span><span>Number of moles of Ne = 11.12/20= 0.556
</span><span>Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3</span><span>
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In order to become a scientific theory the three
categories that it must pass are the following:
1) Can the phenomena be recreated in a laboratory setting?
2) Can variables be changed, yet still result in like observations?
3) Is the phenomena truly natural or was it the result of a man-made force
enacting upon it?
Answer:
a. 123.9°C
b.
c.
Explanation:
Hello, I'm attaching a picture with the numerical development of this exercise.
a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.
b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.
c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.
** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.
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Answer:
The ΔH for the reaction is -456.5 KJ
Explanation:
Here we want to determine ΔH for the reaction;
Mathematically;
ΔH = ΔH(product) - ΔH(reactant)
In the case of the first reaction;
ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)
From the other reactions, we can get the respective ΔH for the individual molecule in the reaction
In second reaction;
Kindly note that for elements, molecule of gases, ΔH = 0
What this means is that throughout the solution;
ΔH(Ca) = 0 KJ
ΔH(O2) = 0 KJ
ΔH(C) = 0 KJ
Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone
So in the second reaction
ΔH = 2ΔH(CaCO3)
Thus;
-2414/2 = ΔH(CaCO3)
ΔH(CaCO3) = -1,207 KJ
Moving to the third reaction, we have;
ΔH = ΔH(CO2)
Hence ΔH(CO2) = -393.5 KJ
For the last reaction;
ΔH = ΔH(CaO)
Hence ΔH(CaO) = -1270 KJ
Going back to equation *
ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)
Using the values of the ΔH of the respective molecules given above,
ΔH = -1270 + (-393.5) - (-1207)
ΔH = -456.5 KJ
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:
