Answer:
C₂Cl₄
Explanation:
To know if free rotation around a bond in a compound is possible, we need to see the structure of the compound (picture in attachment).
In single bonds, which are formed by σ bonds, the atoms are not fixed in a single position, and free rotation is permitted.
Double and triple bonds are formed by a σ bond and one or two π bonds, respectively. These bonds do not allow rotation, since it is not possible to twist the ends without breaking the π bond.
The chloroethylene (C₂Cl₄) has two carbons with an sp2-sp2 hybridization, they are bonded together by a double bond. <u>Free rotation on this bond is not possible, because six atoms, including the carbon atoms, doubly bonded and the four chlorine atoms bonded to them, must be on the same plane. </u>
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer: The metalloids are a group of elements in the periodic table. They are located to the right of the post-transition metals and to the left of the non-metals. Metalloids have some properties in common with metals and some in common with non-metals.
Explanation:
