Answer:
Explanation:
First digit of the 2p^3 gives you value of n, in this case its = 2, So, n= 2
Second alphabet gives you the value of l,
l=0 =s
l=1 =p
l=3=d
l=4=f
since "p" is the alphabet in 2p^3, so in your case lt shoudlbe = 1 right?
ml= -l to +l , that is -1, 0, +1
Ms= +1/2 or -1/2 alaways remains same foe evrything.
Answer:
Fluorine
General Formulas and Concepts:
<u>Chemistry</u>
- Reading a Periodic Table
- Periodic Trends
- Electronegativity - the tendency for an element to attract an electron to itself
- Z-effective and Coulomb's Law, Forces of Attraction
Explanation:
The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.
Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.
However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.
Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.
Answer:
6l
Explanation: convert temperature to kelvin by adding 273 and then input the values into the formula with the given constant
2*v=0.5*0.8206*288 then divide both sides by 2 and get the amount in litres which is 6
We consider the given reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O as applies to an acid-base titration.
moles NaOH = c · V = 0.2767 mmol/mL · 25.34 mL = 7.011578 mmol
moles H2SO4 = 7.011578 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH
= 3.505789 mmol
Hence
[H2SO4]= n/V = 3.505789 mmol / 39.54 mL = 0.08866 M
The answer to this question is [H2SO4] = 0.08866 M
Answer is: the hydrogen ion concentration is 10⁻¹⁰ mol/L, solution is basic.
[OH⁻] =0.00001 mol/L = 10⁻⁴ M.
[OH⁻]·[H⁺<span>] = Kw.
</span>0.00001 mol/L ·[H⁺] = 10⁻¹⁴ mol²/L².
[H⁺] = 10⁻¹⁴ mol²/L²÷ 10⁻⁴ mol/L.
[H⁺] = 10⁻¹⁰ mol/L.
pH = -log[H⁺].
pH = -log(10⁻¹⁰ mol/L).
pH = 10.
If pH is less than seven, solution is acidic; greater than seven, solution is basic; equal seven, solution is neutral.
