Answer:
The mixture of ethanol and water is a liquid-liquid solution. The solvent is Water.
Explanation:
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Answer:</h2>
Valance electrons can be determined by <u>Group</u> on the periodic table
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Explanation:</h2>
- Valence electrons are the electrons present in the outermost shell of an atom. We can determine the total number of valence electrons present in an atom by checking at its Group in which it is placed in the periodic table. For example, atoms in Groups 1 the number of valence electron is one and for group 2 the number of valence electrons is 2.
- The groups have number of valance electrons as follow:
Group 1 - 1 valence electron.
Group 2 - 2 valance electrons.
Group 13 - 3 valence electrons.
Group 14 - 4 valance electrons.
Group 15 - 5 valence electrons.
Group 16 - 6 valence electrons.
Group 17 - 7 valence electrons.
Group 18 - 8 valence electrons.
Result: No of valence electron can be determined by the group no. of the element.
Answer: In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions. Metals are shiny.
Explanation: Hope this helped!
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
