Answer:
1.Hydrogenation of Alkenes and akynes.
2.Reaction of alkylhalides.
3. Halogenation.
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
Please find the structure attached as an image
Explanation:
Based on the characteristics ending name (-ene) of the organic compound above, it belongs to the ALKENE GROUP. Alkenes are characterized by the possession of a carbon to carbon double bond (C=C) in their structure.
- But-3-ene tells us that the organic compound has four straight carbon atoms with the C=C (double bond) located on the THIRD carbon depending on if we count from right to left or vice versa.
- 2 methyl indicates that the methyl group (-CH3) is located as an attachment on the second carbon (carbon 2).
N.B: In the structure attached below, the counting is from the left to right (→).
It is topsoil, subsoil, and parent material
A mutation that gives a rabbit a third ear
