Answer:
ions are surrounded by hydrogen ends with positive partial charge. In this way the salt is dissolved in water. Sugar is a molecular compound formed by covalent bonds. In a polar covalent bond, electrons are shared unevenly
Explanation:
I think your answer is “true”, sorry if I’m wrong and hope this helps
The number<span> of protons in the nucleus of an </span>atom is equal to <span>the </span><span>atomic number of an element. You can also find it by subtracting the number of neutrons from the atomic mass. Atomic Number = Atomic Mass - No. of Neutrons.</span>
The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.
Explanation:
Flow of electrons is possible with the help of a conducting medium like metal wire.
A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.
If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.
Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.
Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation:
