The principal quantum number, <span>nn</span>, designates the principal electron shell. Because n describes the most probable distance of the electrons from the nucleus, the larger the number n is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. n can be any positive integer starting at 1, as <span><span>n=1</span><span>n=1</span></span> designates the first principal shell (the innermost shell). The first principal shell is also called the ground state, or lowest energy state. This explains why <span>nn</span> can not be 0 or any negative integer, because there exists no atoms with zero or a negative amount of energy levels/principal shells. When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where <span><span>n=2</span><span>n=2</span></span>. This is called absorption because the electron is "absorbing" photons, or energy. Known as emission, electrons can also "emit" energy as they jump to lower principle shells, where n decreases by whole numbers. As the energy of the electron increases, so does the principal quantum number, e.g., n = 3 indicates the third principal shell, n = 4 indicates the fourth principal shell, and so on.
Answer: a . 152g/mol b. 102g/mol c. 183g/mol
Explanation:
By stating the atomic masses of each element in the questions, we have;
Fe= 56, S= 32, O= 16, Al = 27, C = 12, H =1 , N = 14, therefore
(a). FeSO4 = 56 + 32 + (16 x 4) = 152g/mol
(b). Al2O3 = (27 x 2) + (16 x 3) = 102g/mol
(c). C7H5NO3S ( Saccharin, an artificial Sweetner) =
(12 x 7) + (1 x 5) + 14 + (16 x 3) + 32 = 183g/mol
I hope this helps answer your question. :)
These problems are very hard to do digitally so I would recommend trying to practice these types of problems on paper :)
Answer:
169.67Ω
Explanation:
This question is asking for the inductive reactance, which is calculated as follows:
X(L) = 2πfL
Where;
X(L) = inductive reactance (Ω)
f = frequency (Hz)
π = 3.142
L = inductance (Henry)
Given the information provided, f = 60Hz, L = 0.450H
X(L) = 2πfL
X(L) = 2 × 3.142 × 60 × 0.450
X(L) = 6.284 × 60 × 0.450
X(L) = 6.284 × 27
X(L) = 169.668
X(L) = 169.67Ω
= 0.15 mol Al₂O₃
