The three isomers of pentane have different structural formulas.
The correct answer is letter <span>C. mixture in which its components retain their identity. A heterogeneous mixture is a mixtures in which the component of the mixed are not uniform. You can see that there are localized regions that have different properties. The components have the capacity to retain their identity.</span>
Answer:
- <u><em>Magnesium and fluorine.</em></u>
Explanation:
<em>Ionic compounds</em> are formed by the electrostatic attraction of cations and anions.
Cations, positive ions, are formed when atoms lose electrons, and anions, negative ions, are formed when atoms gain electrons.
When two different atoms have similar atraction for electrons (electronegativity) they will not donate to nor catch electrons from each other, so cations and anions will not be formed. Instead, the atoms would prefer to share electrons forming covalent bonds to complete their outermost shell (octet rule).
Then, in order to form ionic compounds the electronegativities have to substantially different. This situation does not happen between two nonmetal elements, which nitrogen and sulfur are. Then, you can predict safely that nitrogen and sulfur will not form an ionic compound.
Ionic compounds, then require the electronegativity difference that exist between some metals and nonmetals. Being magnesium an alkaline earth metal, its electronegativity is very low. On the other hand, fluorine the first element of the group 17, has the highest electronegativity of all the elements.Thus magnesium and fluorine will have enough electronegativity difference to justify the exchange of electrons, forming ions and, consequently, ionic compounds.
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:
