What is the electron configuration for helium (He)?A. 1s1
B. 1s2
C. 1s22s1
D. 1s22s2
Answers
The electronic configuration is the system of the electron distribution in the shells of the atom. The electronic configuration for He is 1s² as it is a noble gas. Thus, option B is correct.
What is electronic configuration?
The electronic configuration is the depiction of the electrons in the orbital level of the atoms around the nucleus. The electrons are distributed in the four orbitals namely, s, p, d, and f.
The orbitals have different sub-levels and can accommodate a different number of electrons. The electronic configuration of helium atom with atomic number 2 can be written as, 1s².
The s orbital can accommodate only two electrons in them and helium having only two electrons gets completely filled. Helium being the inert gas has the most stable configuration.
Therefore, 1s² is the electronic configuration of He.
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Answer:
B. 1s2
this is your answer....
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Answer:
Unit rate at which Martin paints in walls per gallon is .
Explanation:
Rate is defined as ratio between two quantities with different units.
Gallons of paint used by Martin = gallons
Portion of wall Martin wishes to pint = wall
Rate at which Martin paints the wall:
Unit rate at which Martin paints in walls per gallon is .
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Final answer:
Protons and neutrons are found in the nucleus of an atom while electrons are found in orbitals surrounding the nucleus. The number of protons determines the type of element the atom is.
Explanation:
In an atom, protons and neutrons are located in the nucleus, which is the central part of the atom. The electrons on the other hand, are situated in the areas around the nucleus called electron shells or orbits. Protons carry a positive charge, neutrons are neutral, and electrons carry a negative charge. The number of protons in an atom is what determines the atomic number and thus, the type of element it is.
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B. A sample of 2.00 g of O2 is removed.
C. A sample of 4.00 g of O2 is added to the 4.80 g of O2 gas in the container.
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Final answer:
By using the ideal gas law and molar mass calculations, the final volumes are found to be A. 65.0 L, B. 8.75 L, and C. 27.5 L.
Explanation:
To calculate the final volume when additional O2 is added or when some O2 is removed, we can use the concept of the molar mass and the ideal gas law that states that volume is directly proportional to the amount of gas, assuming pressure and temperature is constant.
The molar mass of O2 is approximately 32.00 g/mol.
A. 0.500 moles of O2 is added. This equals 0.500 * 32 g = 16 g. The total mass in the system is now 20.8 g. If the original 15.0 L represented 4.80 g, now 20.8 g would represent 15.0 L * 20.8/4.80 = 65.0 L.
B. 2.00 g of O2 is removed. So, the total mass in the system is now 2.80 g. If the original 15.0 L represented 4.80 g, now 2.80 g would represent 15.0 L * 2.80/4.80 = 8.75 L.
C. 4.00 g of O2 is added. So, the total mass in the system is now 8.80 g. If the original 15.0 L represented 4.80 g, now 8.80 g would represent 15.0 L * 8.80/4.80 = 27.5 L.
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Answers
C2H2 + 5/2O2 = 2CO2 + H2O
We are given the amount of C2H2 being burned. This will be our starting point.
64.0 g C2H2 (1 mol C2H2 / 26.02 g C2H2) (1 mol H2O/1mol C2H2) ( 18.02 g H2O/1mol H2O) = 44.32 g H2O.
Thus, the answer is 44.32 g H2O.
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Answer:
CO2,H2S,HALON,FREON that's is the four toxicants that are heavier than air