Explanation: energy levels are three(counting number of cyecles) outer electrons are 3 also
Answer:3,3
Answer:
1. HBr>HCl> H2S >BH3
2.K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
Explanation:
As one goes down a row in the Periodic Table the properties that determine the acid strength can be observed.
The atoms get larger in radius meaning that in strength, the strength of the bonds get weaker, conversely meaning that the acids get stronger.
For the halogen-containing acids above following the rows and periods, HBr has the strongest bond and is the strongest acid and others follow in this order.
HBr>HCl> H2S >BH3
Acid Dissociation Constant provides us with information known as the ionization constant which comes in handy to measure the acid's strength. The meaning of the proportions are thus, the higher the Ka value, the stronger the acid i.e. it liberates more number of hydrogen ions per mole of acid in solution.
In solution strong acids completely dissociate hence, the value of dissociation constant of strong acids is very high.
Following the cues above on Ka;
K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
D. precision. At first glance you can mark out A and B because the answers does not relate to the question. If question had said "what do you call it when the measurement is close to the actual answer" then you would have picked C. So that leaves you D. precision.
Answer:Comparison of Ionic and Covalent Bonds
In an ionic bond, the atoms are bound together by the electrostatic forces in the attraction between ions of opposite charge. ... For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl. In a covalent bond, the atoms bond by sharing electrons.
pls add Brainliest
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:
