Electrons are free to move throughout metal substance, shared throughout so electricity and heat are conducted well
The number of electrons it take to fill the mos formed from the combination of the 3s orbitals of two atoms simply is 14 electrons.
<h3>How electrons are distributed in the 3s orbitals.</h3>
The 3s orbital possess two different spherical nodes which is highly connected with the principal quantum number. In order words, it has 2 radial nodes. Also the shape of the 3s orbital is spherical in shape.
That being said, from the context of the above given task, the number of electrons which fill the mos formed from the combination of the 3s orbitals of two atoms is fourteen electrons.
However, the electron configuration of an atom simply is the arrangement of electrons in the electron shell or orbit of the atom of that element.
In conclusion, it can be deduced from above s orbital has a maximum of two electrons and this energy increases as the orbitals increases.
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Answer: Chemical Reactions
Explanation:
Chemical reactions often involve changes in energy due to the breaking and formation of bonds.
Answer:
B.
Explanation:
When water separates into hydrogen and oxygen gas, a decomposition reaction is taking place (one substance breaking apart or breaking down into two or more substances).
The law of conservation of mass tells us that mass is not created or destroyed during a chemical reaction, so the total mass will remain the same. This is supported by the diagram on the page, which shows that in the reactants, there are four hydrogen atoms and two oxygen atoms, and this amount is the same on the product side.
Answer: 0.0944 gram of H2
Explanation:
Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:
2.0 atm x (973 K / 298 K) = 6.53 atm
Where
Kc = Kp because the moles of product equals the moles of reactants.
At equilibriuim, the amounts are
P(H2) = 6.53 - x
P(CO2) = 6.53 - x
P(H2O) = x
P(CO) = x
Kc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]
Take the square root of each side
(.534)^0.5 = x / (6.53 - x)
x = 0.731 (6.53 - x)
x = 4.77 - 0.731x
1.731x = 4.77
x = 4.77 / 1.731 = 2.76 atm
P(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm
P(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm
PV = nRT
n = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H2
0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g
