It is energetically favorable for all atoms to have a complete outer
electron shell. Loosely, the atoms on the left hand side of the periodic
table only have a few extra electrons in their outer shell so it is
energetically favorable for them to lose them. The atoms on the right
hand side of the periodic table almost have enough electrons in their
outer shell and so they have a tendency to gain them.
Once electrons have left an electron shell, an atom will have a positive
charge because it has more protons (positive charges) than electrons
(negative charges). Similarly, an electron which has gained electrons to
complete its outer shell will have a negative charge because it now has
more electrons (negative charge) than protons (positive charge).
Answer:
b.)
Explanation:
Because you would have a different weight on the moon because of it's low gravity
Answer:
Second element(Titanium); [Ar] 3d2 4s2
Third element(Vanadium):Ar 3d3 4s2
Explanation:
Given that there are only three d orbitals in universe L instead of five, the electronic configuration of the second and third elements in the first transition series will now look thus;
Second element(Titanium); [Ar] 3d2 4s2
Third transition element(Vanadium):Ar 3d3 4s2
Hence, the electronic configuration of Titanium and Vanadium in universe L is just the same as what it is on earth.
Answer:
1 mole of HCl or NaOH gives you 1 mole of H2O , then the number of moles in H2O is: [ 1÷1×1 ] = 1 mole.
Explanation:
The molar mass of the protein is 45095 g/mol.
The mass of a sample of a chemical compound divided by the quantity, or number of moles in the sample, measured in moles, is known as the molar mass of that compound.
The expression of molar mass of protein is
M₂ = (W₂/P) (RT/V)
Given;
W₂ = 1.31g
P = 4.32 torr = 5.75 X 10⁻³ bar
R = 0.083 Lbar/mol/K
T = 25°C = 298.15 K
V = 125 ml = 0.125 L
Putting all the values in the above formula
M₂= (1.31 g/5.75 X 10⁻³ bar) X (0.083 Lbar/mol/K X 2)/0.125 L)
M₂ = 45095 g/mol
Thus, the molar mass of the protein is 45095 g/mol.
Learn more about the Molar mass with the help of the given link:
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