Answer:
Each cell contributes power of 14.7 Watts.
Explanation:
Total power generated by alkali furl cells = 14,800 Watts
Total number of fuel cells = 1,008
Power produced by one fuel cell = x
Total power generated = 
Each cell contributes power of 14.7 Watts.
Answer:
528.25 Calories
Explanation:
From the question given above, the following data were obtained:
Energy in Joules = 2210.20 J
Energy in calories =?
We can convert 2210.20 J to calories by doing the following:
4.184 J = 1 cal
Therefore,
2210.20 J = 2210.20 J × 1 cal / 4.184 J
2210.20 J = 528.25 Calories
Thus, 2210.20 J is equivalent to 528.25 Calories.
Start studying Pressure - Volume Relationships in Gases (Boyle's Law). Learn vocabulary, terms ...Select<span> all that apply. V2 = k/P2 V2 = P1V1/P2 ... What </span>two variables<span> are </span>held constant when testing Boyle's Law in a manometer<span>? Temperature hope this helps
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The element of lowest atomic number whose electron configuration has 4 completely filled p orbitals is, Xenon, Xe with electron configuration:1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶.
According to the question, we are required to identify the element with the lowest atomic number whose electronic configuration has four completely filled p sub-shells.
Evidently, the shell with energy level one has no p orbital.
- Therefore, the existence of p orbitals in electron configuration begins with shell energy level, n=2.
- However, we must know that the p-orbital comprises of 3 sub-orbitals namely;
p(x) , p(y) and p(z) with two electrons of opposing spin in each sub orbital.
- Therefore, the total number of electrons in the p orbital is 6 electrons.
In essence, the element in question should have electron configuration;
- 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶.
The electron configuration above contains 54 electrons and has atomic number, 54.
In essence, the element of lowest atomic number whose electron configuration has 4 completely filled p orbitals is, Xenon, Xe.
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