Maximum number of covalent bonds that an oxygen atom can make with hydrogen is 2.
- the ground state electronic configuration of oxygen is 2s² 2p⁴ that means it has 6 electrons in its valence shell and require two electrons are required to complete its octate.
- Two bonds are created when an electron donor atom shares the two needed electrons with oxygen. The ability of two oxygen atoms to share valence electrons results in the creation of a double bond between the two atoms.
- There are no longer any empty orbitals in the octet of oxygen after it is complete. As a result, it is unable to accept more electrons or create more bonds.
Therefore, Oxygen can only generate two bonds because it needs two additional electrons to complete its octet, after which it will run out of empty orbitals in which to receive additional electrons and create additional bonds.
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Answer:
Explanation:
The given reaction equation is:
2A + 4B → C + 3D
We know the mass of compound A in the reaction above. We are to find the mass of compound D.
We simply work from the known mass to calculate the mass of the unkown compound D
Using the mole concept, we can find the unknown mass.
Procedures
- We first find the molar mass of the compound A from the atomic units of the constituent elements.
- We then use the molar mass of A to calculate its number of moles using the expression below:
Number of moles of A = 
- Using the known number of moles of A, we can work out the number of moles of D.
From the balanced equation of the reaction, it is shown that:
2 moles of compound A was used up to produced 3 moles of D
Then 
x number of moles of A would give the number of moles of D
- Now that we know the number of moles of D, we can find its mass using the expression below:
Mass of D = number of moles of D x molar mass of D
HCl is the limiting reactant, and 3.5 mol H2<span> can be formed</span>
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.
It looks all correct to me, great job!
