Covalent bonds or interactions are overcome when a nonmetal extended network melts.
Typically, nonmetals form covalent bonds with one another. A polyatomic ion's atoms are joined by a form of link called covalent bonding. A covalent bond requires two electrons, one from each of the two atoms that are connecting.
One technique to depict the formation of covalent connections between atoms is with Lewis dot formations. The number of unpaired electrons and the number of bonds that can be formed by each element are typically identical. Each element needs to share an unpaired electron in order to establish a covalent bond.
Therefore, covalent bonds or interactions are overcome when a nonmetal extended network melts.
Learn more about covalent bonds here;
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Answer:
7.146
Explanation:
use the equilibrium equation
Answer: The new concentration of a solution of 
is 0.2 M 10.0 mL of a 2.0 M solution is diluted to 100 mL.
Explanation:
Given: 
= 10.0 mL, 
= 2.0 M
= 100 mL, 
= ?
Formula used to calculate the new concentration is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the new concentration of a solution of is 0.2 M 10.0 mL of a 2.0 M solution is diluted to 100 mL.
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.
Adding a catalyst as this would speed up the reaction and the rest would slow it down
